6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero the velocity of the centre of mass remainsLet us recall that the linear momentum of a constant. (We assume throughout the particle is defined as discussion on systems of particles in this p = m v (6.12) chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F = (6.13) another, the individual particles may have dt Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 101 complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle. The vector Eq. (6.18a) is equivalent to three scalar equations, Px = c1, Py = c2 and Pz = c3 (6.18 b) (a) (b) Here Px, Py and Pz are the components of the total linear momentum vector P along the x–, y– Fig. 6.14 (a) Trajectories of two stars, S1 (dotted line) and z–axes respectively; c1, c2 and c3 are and S2 (solid line) forming a binary constants. system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest. move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b). In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference. (a) (b) In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star Fig. 6.13 (a) A heavy nucleus radium (Ra) splits into moves like a free particle, as shown in Fig.6.14 a lighter nucleus radon (Rn) and an alpha (a). The trajectories of the two stars of equal particle (nucleus of helium atom). The CM mass are also shown in the figure; they look of the system is in uniform motion. complicated. If we go to the centre of mass (b) The same spliting of the heavy nucleus radium (Ra) with the centre of mass at frame, then we find that there the two stars rest. The two product particles fly back are moving in a circle, about the centre of to back. mass, which is at rest. Note that the position of the stars have to be diametrically opposite As an example, let us consider the to each other [Fig. 6.14(b)]. Thus in our frame radioactive decay of a moving unstable particle, of reference, the trajectories of the stars are a combination of (i) uniform motion in a straightlike the nucleus of radium. A radium nucleus line of the centre of mass and (ii) circulardisintegrates into a nucleus of radon and an orbits of the stars about the centre of mass.alpha particle. The forces leading to the decay As can be seen from the two examples,are internal to the system and the external separating the motion of different parts of aforces on the system are negligible. So the total system into motion of the centre of mass andlinear momentum of the system is the same motion about the centre of mass is a verybefore and after decay. The two particles useful technique that helps in understanding produced in the decay, the radon nucleus and the motion of the system. the alpha particle, move in different directions in such a way that their centre of mass moves 6.5 VECTOR PRODUCT OF TWO VECTORS along the same path along which the original decaying radium nucleus was moving We are already familiar with vectors and their [Fig. 6.13(a)]. use in physics. In chapter 5 (Work, Energy, Power) If we observe the decay from the frame of we defined the scalar product of two vectors. An reference in which the centre of mass is at rest, important physical quantity, work, is defined as the motion of the particles involved in the decay a scalar product of two vector quantities, force looks particularly simple; the product particles and displacement. Reprint 2025-26 102 PHYSICS We shall now define another product of two A simpler version of the right hand rule is vectors. This product is a vector. Two important the following : Open up your right hand palm quantities in the study of rotational motion, and curl the fingers pointing from a to b. Your namely, moment of a force and angular stretched thumb points in the direction of c. momentum, are defined as vector products. It should be remembered that there are two angles between any two vectors a and b . In Definition of Vector Product Fig. 6.15 (a) or (b) they correspond to θ(as shown) A vector product of two vectors a and b is a and (3600– θ). While applying either of the above vector c such that rules, the rotation should be taken through the (i) magnitude of c = c = ab sinθ where a and b smaller angle (<1800) between a and b. It is θ are magnitudes of a and b and θ is the here. angle between the two vectors. Because of the cross (×) used to denote the (ii) c is perpendicular to the plane containing vector product, it is also referred to as cross product. a and b. • Note that scalar product of two vectors is (iii) if we take a right handed screw with its head commutative as said earlier, a.b = b.a lying in the plane of a and b and the screw The vector product, however, is not perpendicular to this plane, and if we turn commutative, i.e. a × b ≠ b × a the head in the direction from a to b, then The magnitude of both a × b and b × a is the the tip of the screw advances in the direction same ( ab sin θ ); also, both of them are of c. This right handed screw rule is perpendicular to the plane of a and b. But the illustrated in Fig. 6.15a. rotation of the right-handed screw in case of Alternately, if one curls up the fingers of a × b is from a to b, whereas in case of b × a it right hand around a line perpendicular to the is from b to a. This means the two vectors are plane of the vectors a and b and if the fingers in opposite directions. We have are curled up in the direction from a to b, then a × b = − b × a the stretched thumb points in the direction of • Another interesting property of a vector c, as shown in Fig. 6.15b. product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have x →− x , y →−y and z →− z . As a result all the components of a vector change sign and thus a →−a , b →−b . What happens to a × b under reflection? a × b →−( a ) × ( − b ) = a × b Thus, a × b does not change sign under reflection. • Both scalar and vector products are distributive with respect to vector addition. Thus, a.( b + c ) = a.b + a.c a × ( b + c ) = a × b + a × c (a) (b) • We may write c = a × b in the component form. For this we first need to obtain some elementary cross products: Fig. 6.15 (a) Rule of the right handed screw for (i) a × a = 0 (0 is a null vector, i.e. a vector defining the direction of the vector with zero magnitude) product of two vectors. This follows since magnitude of a × a is (b) Rule of the right hand for defining the direction of the vector product. a 2 sin0° = 0 . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 103 From this follow the results ˆ ˆ ˆ i j k (i) ˆi × ˆi = 0, ˆj × ˆj = 0, kˆ × kˆ = 0 a × b = 3 − 4 5 = 7 ˆi − ˆj − 5 kˆ (ii) ˆi × ˆj = kˆ − 2 1 − 3 Note that the magnitude of ˆi × ˆj is sin900 Note b × a = −7ˆi + ˆj + 5 kˆ ⊳ or 1, since ˆi and ˆj both have unit magnitude and the angle between them is 900. 6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY Thus, ˆi × ˆj is a unit vector. A unit vector In this section we shall study what is angular perpendicular to the plane of ˆi and ˆj and velocity and its role in rotational motion. We related to them by the right hand screw rule is have seen that every particle of a rotating body moves in a circle. The linear velocity of the ˆk . Hence, the above result. You may verify particle is related to the angular velocity. The similarly, relation between these two quantities involves ˆ j × kˆ = ˆi and kˆ × ˆi = ˆj a vector product which we learnt about in the last section. From the rule for commutation of the cross Let us go back to Fig. 6.4. As said above, inproduct, it follows: rotational motion of a rigid body about a fixed ˆ j × ˆi = − kˆ , kˆ × ˆj = − ˆi, ˆi × kˆ = − ˆj axis, every particle of the body moves in a circle, Note if ˆi, ˆj, kˆ occur cyclically in the above vector product relation, the vector product is positive. If ˆi, ˆj, kˆ do not occur in cyclic order, the vector product is negative. Now, a × b = (a x ˆi + a y ˆj + a z kˆ ) × (b x ˆi + b y ˆj + b z kˆ ) = a x b y kˆ − a x b z ˆj − a y b x kˆ + a y b z ˆi + a z b x ˆj − a z b y ˆi = (a y b z − a z b y )i + (a z b x − a x b z ) j + (a x b y − a y b x )k We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember. ˆ ˆ ˆ i j k a × b = a x a y a z b x b y b z u Example 6.4 Find the scalar and vector Fig. 6.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed products of two vectors. a = (z-) axis moves in a circle with centre (C) and b = on the axis.) Answer which lies in a plane perpendicular to the axis a i b = (3ˆi − 4 ˆj + 5 kˆ )i( − 2ˆi + ˆj − 3 kˆ ) and has its centre on the axis. In Fig. 6.16 we redraw Fig. 6.4, showing a typical particle (at a = −6 − 4 − 15 point P) of the rigid body rotating about a fixed = −25 axis (taken as the z-axis). The particle describes Reprint 2025-26 104 PHYSICS a circle with a centre C on the axis. The radius and points out in the direction in which a right of the circle is r, the perpendicular distance of handed screw would advance, if the head of the the point P from the axis. We also show the screw is rotated with the body. (See Fig. 6.17a). linear velocity vector v of the particle at P. It is The magnitude of this vector is ω = d θ dt along the tangent at P to the circle. referred as above. Let P′ be the position of the particle after an interval of time ∆t (Fig. 6.16). The angle PCP′ describes the angular displacement ∆θ of the particle in time ∆t. The average angular velocity of the particle over the interval ∆t is ∆θ/∆t. As ∆t tends to zero (i.e. takes smaller and smaller values), the ratio ∆θ/∆t approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocity by ω (the Greek letter omega). We know from our study Fig. 6.17 (a) If the head of a right handed screw of circular motion that the magnitude of linear rotates with the body, the screw velocity v of a particle moving in a circle is advances in the direction of the angular related to the angular velocity of the particle ω velocity ω. If the sense (clockwise or by the simple relation υ = ωr , where r is the anticlockwise) of rotation of the body changes, so does the direction of ω.radius of the circle. We observe that at any given instant the relation v = ωr applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by v i = ωri (6.19) The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, r = 0 , and hence v = ω r = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed. Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body. We have characterised pure translation of a body by all parts of the body having the same Fig. 6.17 (b) The angular velocity vector ω is directed velocity at any instant of time. Similarly, we along the fixed axis as shown. The linear may characterise pure rotation by all parts of velocity of the particle at P is v = ω × r. the body having the same angular velocity at It is perpendicular to both ωωωωω and r and any instant of time. Note that this is directed along the tangent to the circle described by the particle. characterisation of the rotation of a rigid body about a fixed axis is just another way of saying We shall now look at what the vector as in Sec. 6.1 that each particle of the body moves product ω × r corresponds to. Refer to Fig. in a circle, which lies in a plane perpendicular 6.17(b) which is a part of Fig. 6.16 reproduced to the axis and has the centre on the axis. to show the path of the particle P. The figure In our discussion so far the angular velocity shows the vector ω directed along the fixed (z–) appears to be a scalar. In fact, it is a vector. We axis and also the position vector r = OP of the shall not justify this fact, but we shall accept particle at P of the rigid body with respect to it. For rotation about a fixed axis, the angular the origin O. Note that the origin is chosen to velocity vector lies along the axis of rotation, be on the axis of rotation. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 105 Now ω × r = ω × OP = ω × (OC + CP) If the axis of rotation is fixed, the direction But ω × OC = 00000 as ωωωωω is along OC of ωωωωω and hence, that of α is fixed. In this case Hence ω × r = ω × CP the vector equation reduces to a scalar equation dω α = (6.22) The vector ω × CP is perpendicular to ω, i.e. dt to the z-axis and also to CP, the radius of the circle described by the particle at P. It is 6.7 TORQUE AND ANGULAR MOMENTUM therefore, along the tangent to the circle at P. In this section, we shall acquaint ourselves with Also, the magnitude of ω × CP is ω (CP) since two physical quantities (torque and angular ω and CP are perpendicular to each other. We momentum) which are defined as vector products shall denote CP by ⊥r and not by r, as we did of two vectors. These as we shall see, are earlier. especially important in the discussion of motion Thus, ω × r is a vector of magnitude ωr⊥ of systems of particles, particularly rigid bodies. and is along the tangent to the circle described by the particle at P. The linear velocity vector v 6.7.1 Moment of force (Torque) at P has the same magnitude and direction. We have learnt that the motion of a rigid body, Thus, in general, is a combination of rotation and v = ωωωωω × r (6.20) translation. If the body is fixed at a point or along In fact, the relation, Eq. (6.20), holds good a line, it has only rotational motion. We know even for rotation of a rigid body with one point that force is needed to change the translationalfixed, such as the rotation of the top [Fig. 6.6(a)]. In this case r represents the position vector of state of a body, i.e. to produce linear the particle with respect to the fixed point taken acceleration. We may then ask, what is the as the origin. analogue of force in the case of rotational We note that for rotation about a fixed motion? To look into the question in a concrete axis, the direction of the vector ω does not situation let us take the example of opening or change with time. Its magnitude may, closing of a door. A door is a rigid body which however, change from instant to instant. For can rotate about a fixed vertical axis passing the more general rotation, both the magnitude and the direction of ωωωωω may change through the hinges. What makes the door from instant to instant. rotate? It is clear that unless a force is applied the door does not rotate. But any force does not 6.6.1 Angular acceleration do the job. A force applied to the hinge line You may have noticed that we are developing cannot produce any rotation at all, whereas a the study of rotational motion along the lines force of given magnitude applied at right angles of the study of translational motion with which to the door at its outer edge is most effective in we are already familiar. Analogous to the kinetic producing rotation. It is not the force alone, but variables of linear displacement (s) and velocity how and where the force is applied is important (v) in translational motion, we have angular in rotational motion. displacement (θ) and angular velocity (ω) in The rotational analogue of force in linear rotational motion. It is then natural to define motion is moment of force. It is also referred to in rotational motion the concept of angular as torque or couple. (We shall use the words acceleration in analogy with linear acceleration moment of force and torque interchangeably.) defined as the time rate of change of velocity in We shall first define the moment of force for the translational motion. We define angular special case of a single particle. Later on we acceleration α as the time rate of change of shall extend the concept to systems of particles angular velocity. Thus, including rigid bodies. We shall also relate it to d ω a change in the state of rotational motion, i.e. is α = (6.21) dt angular acceleration of a rigid body. Reprint 2025-26 106 PHYSICS of the line of action of F from the origin and F⊥=( F sin θ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin. One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same. 6.7.2 Angular momentum of a particle Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue Fig. 6.18 τττττ ===== r × F, τττττ is perpendicular to the plane of linear momentum. We shall first define containing r and F, and its direction is angular momentum for the special case of a given by the right handed screw rule. single particle and look at its usefulness in the context of single particle motion. We shall then If a force acts on a single particle at a point extend the definition of angular momentum to P whose position with respect to the origin O is systems of particles including rigid bodies. given by the position vector r (Fig. 6.18), the Like moment of a force, angular momentum moment of the force acting on the particle with is also a vector product. It could also be referred respect to the origin O is defined as the vector to as moment of (linear) momentum. From this product term one could guess how angular momentum τ = r × F (6.23) is defined. The moment of force (or torque) is a vector Consider a particle of mass m and linear quantity. The symbol τττττ stands for the Greek momentum p at a position r relative to the origin letter tau. The magnitude of τττττ is O. The angular momentum l of the particle with τ = r F sinθ (6.24a) respect to the origin O is defined to be l = r × p (6.25a)where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force The magnitude of the angular momentum F and θ is the angle between r and F as vector is shown. l = r p sinθ (6.26a) Moment of force has dimensions M L2 T -2. where p is the magnitude of p and θ is the angle Its dimensions are the same as those of work between r and p. We may write or energy. It is, however, a very different physical l = r p⊥ or r ⊥ p (6.26b)quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of where r⊥ (= r sinθ) is the perpendicular distance moment of force is newton metre (N m). The of the directional line of p from the origin and magnitude of the moment of force may be p ⊥=( p sin θ) is the component of p in a directionwritten perpendicular to r. We expect the angular τ = (r sin θ)F = r⊥ F (6.24b) momentum to be zero (l = 0), if the linear or τ = r F sin θ = rF ⊥ (6.24c) momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p where r⊥ = r sinθ is the perpendicular distance passes through the origin θ = 00 or 1800. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 107 The physical quantities, moment of a force and angular momentum, have an important An experiment with the bicycle rim relation between them. It is the rotational Take a analogue of the relation between force and linear bicycle rim momentum. For deriving the relation in the and extend context of a single particle, we differentiate its axle on l = r × p with respect to time, both sides. Tie two d l d = ( r × p ) s t r i n g s d t d t at both ends Applying the product rule for differentiation A and B, to the right hand side, as shown in the d d r d p ( r × p ) = × p + r × a d j o i n i n g d t d t d t figure. Hold Now, the velocity of the particle is v = dr/dt both the and p = m v Initially After s t r i n g s together in dr one hand such that the rim is vertical. If you Because of this × p = v × m v = 0, dt leave one string, the rim will tilt. Now keeping the rim in vertical position with both the stringsas the vector product of two parallel vectors in one hand, put the wheel in fast rotation vanishes. Further, since dp / dt = F, around the axle with the other hand. Then leave d p one string, say B, from your hand, and observe r × = r × F = t dt what happens. The rim keeps rotating in a vertical plane d and the plane of rotation turns around the string Hence ( r × p ) = τ A which you are holding. We say that the axis dt of rotation of the rim or equivalently or (6.27) its angular momentum precesses about the string A. Thus, the time rate of change of the angular The rotating rim gives rise to an angular momentum of a particle is equal to the torque momentum. Determine the direction of this acting on it. This is the rotational analogue of angular momentum. When you are holding the the equation F = dp/dt, which expresses rotating rim with string A, a torque is generated. (We leave it to you to find out how the torque isNewton’s second law for the translational motion generated and what its direction is.) The effect of a single particle. of the torque on the angular momentum is to make it precess around an axis perpendicular Torque and angular momentum for a system to both the angular momentum and the torque. of particles Verify all these statements. To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles, particle has mass mi and velocity vi) We may write the total angular momentum of a system of particles as (6.25b) The angular momentum of the ith particle is given by li = ri × pi This is a generalisation of the definition of angular momentum (Eq. 6.25a) for a singlewhere ri is the position vector of the ith particle particle to a system of particles.with respect to a given origin and p = (mivi) is Using Eqs. (6.23) and (6.25b), we getthe linear momentum of the particle. (The Reprint 2025-26 108 PHYSICS d L d d l Note that like Eq.(6.17), Eq.(6.28b) holds = ( l ) = ∑ = ∑ τ (6.28a) good for any system of particles, whether it is a d t d t i d t i rigid body or its individual particles have all where τi is the torque acting on the ith particle; kinds of internal motion. τi = ri × Fi Conservation of angular momentum The force Fi on the ith particle is the vector ext If τext = 0, Eq. (6.28b) reduces to Fi sum of external forces acting on the particle d L = 0 and the internal forces iFint exerted on it by the dt other particles of the system. We may therefore or L = constant. (6.29a) separate the contribution of the external and Thus, if the total external torque on a system the internal forces to the total torque of particles is zero, then the total angular momentum of the system is conserved, i.e. τ = ∑ τ i = ∑ ri × Fi as remains constant. Eq. (6.29a) is equivalent to i i three scalar equations, τ = τext + τ int , Lx = K1, Ly = K2 and Lz = K3 (6.29 b) Here K1, K2 and K3 are constants; Lx, Ly and τ ext = ∑ri × Fi ext Lz are the components of the total angular where i momentum vector L along the x,y and z axes respectively. The statement that the total i × Fiint τ int = ∑r and angular momentum is conserved means that i each of these three components is conserved. We shall assume not only Newton’s third law Eq. (6.29a) is the rotational analogue of of motion, i.e. the forces between any two particles Eq. (6.18a), i.e. the conservation law of the total of the system are equal and opposite, but also that linear momentum for a system of particles. these forces are directed along the line joining the Like Eq. (6.18a), it has applications in many two particles. In this case the contribution of the practical situations. We shall look at a few of internal forces to the total torque on the system is the interesting applications later on in zero, since the torque resulting from each action- this chapter. reaction pair of forces is zero. We thus have, τint = 0 and therefore τ = τττext.ττ u Example 6.5 Find the torque of a force Since τ = ∑ τ i , it follows from Eq. (6.28a) + – about the origin. The force acts on a particle whose position vector is .that d L = τ ext (6.28 b) Answer Here r = ˆi − ˆj + kˆ d t and F = 7 ˆi + 3 ˆj − 5 kˆ . Thus, the time rate of the total angular We shall use the determinant rule to find themomentum of a system of particles about a τ = r × Fpoint (taken as the origin of our frame of torque reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (6.28 b) is the generalisation of the single particle case of Eq. (6.23) to a system of particles. Note that when we have only one or ⊳ particle, there are no internal forces or torques. Eq.(6.28 b) is the rotational analogue of Example 6.6 Show that the angular u momentum about any point of a single d P = Fext (6.17) particle moving with constant velocity d t remains constant throughout the motion. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 109 Answer Let the particle with velocity v be at acceleration nor angular acceleration. This means point P at some instant t. We want to calculate (1) the total force, i.e. the vector sum of the the angular momentum of the particle about an forces, on the rigid body is zero; arbitrary point O. n F1 + F2 + ... + Fn = =∑i 1 Fi = 0 (6.30a) If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body. (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero, n τ + τ i = 0 (6.30b) 1 2 + ... + τ n = =∑i 1 τ Fig 6.19 If the total torque on the rigid body is zero, The angular momentum is l = r × mv. Its the total angular momentum of the body does magnitude is mvr sinθ, where θ is the angle not change with time. Eq. (6.30 b) gives the between r and v as shown in Fig. 6.19. Although condition for the rotational equilibrium of the the particle changes position with time, the line body. of direction of v remains the same and hence One may raise a question, whether theOM = r sin θ. is a constant. rotational equilibrium condition [Eq. 6.30(b)] Further, the direction of l is perpendicular remains valid, if the origin with respect to whichto the plane of r and v. It is into the page of the the torques are taken is shifted. One can showfigure.This direction does not change with time. Thus, l remains the same in magnitude and that if the translational equilibrium condition direction and is therefore conserved. Is there [Eq. 6.30(a)] holds for a rigid body, then such a any external torque on the particle? ⊳ shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the 6.8 EQUILIBRIUM OF A RIGID BODY location of the origin about which the torques are taken. Example 6.7 gives a proof of this result We are now going to concentrate on the motion in a special case of a couple, i.e. two forcesof rigid bodies rather than on the motion of acting on a rigid body in translationalgeneral systems of particles. equilibrium. The generalisation of this result to We shall recapitulate what effect the external forces have on a rigid body. (Henceforth n forces is left as an exercise. we shall omit the adjective ‘external’ because Eq. (6.30a) and Eq. (6.30b), both, are vector unless stated otherwise, we shall deal with only equations. They are equivalent to three scalar external forces and torques.) The forces change equations each. Eq. (6.30a) corresponds to the translational state of the motion of the rigid n n n body, i.e. they change its total linear =∑i 1 Fix = 0 , =∑i 1 Fiy = 0 and =∑i 1 Fiz = 0 (6.31a)momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The where Fix, Fiy and Fiz are respectively the x, y and total torque on the body may not vanish. Such z components of the forces Fi. Similarly, Eq. a torque changes the rotational state of motion (6.30b) is equivalent to three scalar equations of the rigid body, i.e. it changes the total angular n n 0momentum of the body in accordance with τix = 0 , τiy = and (6.31b) =∑i 1 =∑i 1 Eq. (6.28 b). where τix, τiy and τiz are respectively the x, y and A rigid body is said to be in mechanical z components of the torque τi .equilibrium, if both its linear momentum and Eq. (6.31a) and (6.31b) give six independentangular momentum are not changing with time, conditions to be satisfied for mechanicalor equivalently, the body has neither linear Reprint 2025-26 110 PHYSICS equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis Fig. 6.20 (b) perpendicular to the plane of the forces must be zero. The force at B in Fig. 6.20(a) is reversed in Fig. 6.20(b). Thus, we have the same rod with The conditions of equilibrium of a rigid body two forces of equal magnitude but acting inmay be compared with those for a particle, which opposite diretions applied perpendicular to the we considered in earlier chapters. Since rod, one at end A and the other at end B. Here consideration of rotational motion does not apply the moments of both the forces are equal, but to a particle, only the conditions for translational they are not opposite; they act in the same sense equilibrium (Eq. 6.30 a) apply to a particle. Thus, and cause anticlockwise rotation of the rod. The for equilibrium of a particle the vector sum of total force on the body is zero; so the body is in all the forces on it must be zero. Since all these translational equilibrium; but it is not in forces act on the single particle, they must be rotational equilibrium. Although the rod is not fixed in any way, it undergoes pure rotation (i.e.concurrent. Equilibrium under concurrent rotation without translation).forces was discussed in the earlier chapters. A pair of forces of equal magnitude but acting A body may be in partial equilibrium, i.e., it in opposite directions with different lines of may be in translational equilibrium and not in action is known as a couple or torque. A couple rotational equilibrium, or it may be in rotational produces rotation without translation. equilibrium and not in translational When we open the lid of a bottle by turning equilibrium. it, our fingers are applying a couple to the lid Consider a light (i.e. of negligible mass) rod [Fig. 6.21(a)]. Another known example is a compass needle in the earth’s magnetic field as(AB) as shown in Fig. 6.20(a). At the two ends (A shown in the Fig. 6.21(b). The earth’s magneticand B) of which two parallel forces, both equal field exerts equal forces on the north and southin magnitude and acting along same direction poles. The force on the North Pole is towards are applied perpendicular to the rod. the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field. Fig. 6.20 (a) Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF ), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational fingers apply a couple to turnequilibrium; F ≠ 0 Fig. 6.21(a) Our ∑ the lid. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 111 length. This point is called the fulcrum. A see- saw on the children’s playground is a typical example of a lever. Two forces F1 and F2, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances d1 and d2 respectively from the fulcrum as shown in Fig. 6.23. Fig. 6.21(b) The Earth’s magnetic field exerts equal and opposite forces on the poles of a Fig. 6.23 compass needle. These two forces form a couple. The lever is a system in mechanical equilibrium. Let R be the reaction of the supportu Example 6.7 Show that moment of a at the fulcrum; R is directed opposite to the couple does not depend on the point about forces F1 and F2. For translational equilibrium, which you take the moments. R – F1 – F2 = 0 (i) Answer For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero, d1F1 – d2F2 = 0 (ii) Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum. Fig. 6.22 In the case of the lever force F1 is usually Consider a couple as shown in Fig. 6.22 some weight to be lifted. It is called the load and acting on a rigid body. The forces F and -F act its distance from the fulcrum d1 is called the respectively at points B and A. These points have load arm. Force F2 is the effort applied to lift the position vectors r1 and r2 with respect to origin load; distance d2 of the effort from the fulcrum O. Let us take the moments of the forces about is the effort arm. the origin. Eq. (ii) can be written as The moment of the couple = sum of the d1F1 = d2 F2 (6.32a) moments of the two forces making the couple or load arm × load = effort arm × effort = r1 × (–F) + r2 × F The above equation expresses the principle = r2 × F – r1 × F of moments for a lever. Incidentally the ratio = (r2–r1) × F F1/F2 is called the Mechanical Advantage (M.A.); But r1 + AB = r2, and hence AB = r2 – r1. F1 d 2 The moment of the couple, therefore, is M.A. = = (6.32b) F2 d1AB × F. Clearly this is independent of the origin, the If the effort arm d2 is larger than the load point about which we took the moments of the arm, the mechanical advantage is greater than forces. ⊳ one. Mechanical advantage greater than one means that a small effort can be used to lift a 6.8.1 Principle of moments large load. There are several examples of a lever An ideal lever is essentially a light (i.e. of around you besides the see-saw. The beam of a negligible mass) rod pivoted at a point along its balance is a lever. Try to find more such Reprint 2025-26 112 PHYSICS examples and identify the fulcrum, the effort and The CG of the cardboard is so located that effort arm, and the load and the load arm of the the total torque on it due to the forces m1g, m2g lever in each case. …. etc. is zero. You may easily show that the principle of If ri is the position vector of the ith particle moment holds even when the parallel forces F1 of an extended body with respect to its CG, then and F2 are not perpendicular, but act at some the torque about the CG, due to the force of angle, to the lever. gravity on the particle is τi = ri × mi g. The total gravitational torque about the CG is zero, i.e. 6.8.2 Centre of gravity i × m i g = 0 (6.33) τ g = ∑ τ i = ∑r Many of you may have the experience of We may therefore, define the CG of a body balancing your notebook on the tip of a finger. as that point where the total gravitational torque Figure 6.24 illustrates a similar experiment that on the body is zero. you can easily perform. Take an irregular- We notice that in Eq. (6.33), g is the same shaped cardboard having mass M and a narrow for all particles, and hence it comes out of the tipped object like a pencil. You can locate by trial summation. This gives, since g is non-zero, and error a point G on the cardboard where it ir = 0. Remember that the position vectorscan be balanced on the tip of the pencil. (The ∑mi cardboard remains horizontal in this position.) (ri) are taken with respect to the CG. Now, in This point of balance is the centre of gravity (CG) accordance with the reasoning given below of the cardboard. The tip of the pencil provides Eq. (6.4a) in Sec. 6.2, if the sum is zero, the origin a vertically upward force due to which the must be the centre of mass of the body. Thus, cardboard is in mechanical equilibrium. As the centre of gravity of the body coincides with shown in the Fig. 6.24, the reaction of the tip is the centre of mass in uniform gravity or gravity- equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the card board due to the forces of gravity like m1g, m2g …. etc, acting on the individual particles that make up the cardboard. Fig. 6.25 Determining the centre of gravity of a body Fig. 6.24 Balancing a cardboard on the tip of a of irregular shape. The centre of gravity G pencil. The point of support, G, is the lies on the vertical AA1 through the point centre of gravity. of suspension of the body A. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 113 free space. We note that this is true because = 30 cm, PG = 5 cm, AK1= BK2 = 10 cm and K1G = the body being small, g does not K2G = 25 cm. Also, W= weight of the rod = 4.00 vary from one point of the body to the other. If kg and W 1= suspended load = 6.00 kg; the body is so extended that g varies from part R1 and R2 are the normal reactions of the to part of the body, then the centre of gravity support at the knife edges. and centre of mass will not coincide. Basically, For translational equilibrium of the rod, the two are different concepts. The centre of R1+R2 –W1 –W = 0 (i) mass has nothing to do with gravity. It depends Note W1 and W act vertically down and R1 only on the distribution of mass of the body. and R2 act vertically up. In Sec. 6.2 we found out the position of the For considering rotational equilibrium, we centre of mass of several regular, homogeneous take moments of the forces. A convenient point objects. Obviously the method used there gives to take moments about is G. The moments of us also the centre of gravity of these bodies, if R2 and W1 are anticlockwise (+ve), whereas the they are small enough. moment of R1 is clockwise (-ve). Figure 6.25 illustrates another way of For rotational equilibrium, determining the CG of an irregular shaped body –R1 (K1G) + W1 (PG) + R2 (K2G) = 0 (ii) like a cardboard. If you suspend the body from It is given that W = 4.00g N and W1 = 6.00g some point like A, the vertical line through A N, where g = acceleration due to gravity. We passes through the CG. We mark the vertical take g = 9.8 m/s2. AA1. We then suspend the body through other With numerical values inserted, from (i) points like B and C. The intersection of the R1 + R2 – 4.00g – 6.00g = 0 verticals gives the CG. Explain why the method or R1 + R2 = 10.00g N (iii) works. Since the body is small enough, the = 98.00 N method allows us to determine also its centre From (ii), – 0.25 R1 + 0.05 W1 + 0.25 R2 = 0 of mass. or R1 – R2 = 1.2g N = 11.76 N (iv) From (iii) and (iv), R1 = 54.88 N, u Example 6.8 A metal bar 70 cm long and R2 = 43.12 N 4.00 kg in mass supported on two knife- Thus the reactions of the support are about edges placed 10 cm from each end. A 6.00 55 N at K1 and 43 N at K2. ⊳ kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. u Example 6.9 A 3m long ladder weighing (Assume the bar to be of uniform cross 20 kg leans on a frictionless wall. Its feet section and homogeneous.) rest on the floor 1 m from the wall as shown in Fig.6.27. Find the reaction forces of the Answer wall and the floor. Answer Fig. 6.26 Figure 6.26 shows the rod AB, the positions of the knife edges K1 and K2 , the centre of gravity of the rod at G and the suspended load at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP Fig. 6.27 Reprint 2025-26 114 PHYSICS The ladder AB is 3 m long, its foot A is at from the axis, the linear velocity is υi = ir ω. The distance AC = 1 m from the wall. From kinetic energy of motion of this particle is Pythagoras theorem, BC = 2 2 m. The forces 1 2 1 2 2 on the ladder are its weight W acting at its centre k i = m i υi = m i ri ω 2 2 of gravity D, reaction forces F1 and F2 of the wall where mi is the mass of the particle. The totaland the floor respectively. Force F1 is kinetic energy K of the body is then given byperpendicular to the wall, since the wall is the sum of the kinetic energies of individualfrictionless. Force F2 is resolved into two particles,components, the normal reaction N and the force of friction F. Note that F prevents the ladder n 1 n 2 2 from sliding away from the wall and is therefore K = ∑ k i = ∑ (m i ri ω ) i =1 2 i =1 directed toward the wall. For translational equilibrium, taking the Here n is the number of particles in the body. forces in the vertical direction, Note ωis the same for all particles. Hence, taking N – W = 0 (i) ω out of the sum, Taking the forces in the horizontal direction, n 1 2 2 i ri ) F – F1 = 0 (ii) K = 2 ω ( ∑i =1 m For rotational equilibrium, taking the We define a new parameter characterisingmoments of the forces about A, the rigid body, called the moment of inertia I , 2 2 F1 −(1/2) W = 0 (iii) given by Now W = 20 g = 20 × 9.8 N = 196.0 N n 2 I = ∑ m i ri (6.34)From (i) N = 196.0 N i =1 With this definition,From (iii) F1 = W 4 2 = 196.0/4 2 = 34.6 N 1 2 From (ii) F = F1 = 34.6 N K = Iω (6.35) 2 2 2 Note that the parameter I is independent of F2 = F + N = 199.0 N the magnitude of the angular velocity. It is a The force F2 makes an angle α with the characteristic of the rigid body and the axis horizontal, about which it rotates. −1 Compare Eq. (6.35) for the kinetic energy oftan α = N F = 4 2 , α = tan (4 2) ≈ 80 ⊳ a rotating body with the expression for the kinetic energy of a body in linear (translational)6.9 MOMENT OF INERTIA motion, We have already mentioned that we are 1 2developing the study of rotational motion parallel K = m υ 2to the study of translational motion with which Here, m is the mass of the body and v is itswe are familiar. We have yet to answer one major velocity. We have already noted the analogy question in this connection. What is the between angular velocity ω (in respect of analogue of mass in rotational motion? We shall rotational motion about a fixed axis) and linear attempt to answer this question in the present velocity v (in respect of linear motion). It is then section. To keep the discussion simple, we shall evident that the parameter, moment of inertia consider rotation about a fixed axis only. Let us I, is the desired rotational analogue of mass in try to get an expression for the kinetic energy of linear motion. In rotation (about a fixed axis), a rotating body. We know that for a body rotating the moment of inertia plays a similar role as about a fixed axis, each particle of the body moves mass does in linear motion. We now apply the definition Eq. (6.34), toin a circle with linear velocity given by Eq. (6.19). calculate the moment of inertia in two simple cases.(Refer to Fig. 6.16). For a particle at a distance Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 115 (a) Consider a thin ring of radius R and mass change in its rotational motion, it can be M, rotating in its own plane around its centre regarded as a measure of rotational inertia of with angular velocity ω. Each mass element the body; it is a measure of the way in which of the ring is at a distance R from the axis, different parts of the body are distributed at and moves with a speed Rω. The kinetic different distances from the axis. Unlike the energy is therefore, mass of a body, the moment of inertia is not a fixed quantity but depends on distribution of 1 2 1 2 2 K = M υ = MR ω mass about the axis of rotation, and the 2 2 orientation and position of the axis of rotation Comparing with Eq. (6.35) we get I = MR 2 with respect to the body as a whole. As a for the ring. measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body. Notice from the Table 6.1 that in all cases, we can write I = Mk2, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, k 2 = L2 12, i.e. k = L 12 . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of Fig. 6.28 A light rod of length l with a pair of gyration. The radius of gyration of a body masses rotating about an axis through about an axis may be defined as the distance the centre of mass of the system and perpendicular to the rod. The total mass from the axis of a mass point whose mass is of the system is M. equal to the mass of the whole body and whose moment of inertia is equal to the moment of (b) Next, take a rigid rod of negligible mass of inertia of the body about the axis. length of length l with a pair of small masses, Thus, the moment of inertia of a rigid body rotating about an axis through the centre of depends on the mass of the body, its shape and mass perpendicular to the rod (Fig. 6.28). size; distribution of mass about the axis of Each mass M/2 is at a distance l/2 from rotation, and the position and orientation of the the axis. The moment of inertia of the masses axis of rotation. is therefore given by From the definition, Eq. (6.34), we can infer (M/2) (l/2)2 + (M/2)(l/2)2 that the dimensions of moments of inertia are Thus, for the pair of masses, rotating about ML2 and its SI units are kg m2. the axis through the centre of mass The property of this extremely important perpendicular to the rod 2 quantity I, as a measure of rotational inertia of I = Ml / 4 the body, has been put to a great practical use. Table 6.1 simply gives the moment of inertia of The machines, such as steam engine and thevarious familiar regular shaped bodies about automobile engine, etc., that produce rotationalspecific axes. (The derivations of these motion have a disc with a large moment ofexpressions are beyond the scope of this inertia, called a flywheel. Because of its largetextbook and you will study them in higher classes.) moment of inertia, the flywheel resists the As the mass of a body resists a change in its sudden increase or decrease of the speed of the state of linear motion, it is a measure of its inertia vehicle. It allows a gradual change in the speed in linear motion. Similarly, as the moment of and prevents jerky motions, thereby ensuring inertia about a given axis of rotation resists a a smooth ride for the passengers on the vehicle. Reprint 2025-26 116 PHYSICS Table 6.1 Moments of inertia of some regular shaped bodies about specific axes Z Body Axis Figure I (1) Thin circular Perpendicular to M R 2 ring, radius R plane, at centre (2) Thin circular Diameter M R2/2 ring, radius R (3) Thin rod, Perpendicular to M L2/12 length L rod, at mid point (4) Circular disc, Perpendicular to M R2/2 radius R disc at centre (5) Circular disc, Diameter M R2/4 radius R (6) Hollow cylinder, Axis of cylinder M R2 radius R (7) Solid cylinder, Axis of cylinder M R2/2 radius R (8) Solid sphere, Diameter 2 M R2/5 radius R 6.10 KINEMATICS OF ROTATIONAL MOTION translation. We wish to take this analogy further. ABOUT A FIXED AXIS In doing so we shall restrict the discussion only We have already indicated the analogy between to rotation about fixed axis. This case of motion rotational motion and translational motion. For involves only one degree of freedom, i.e., needs example, the angular velocity ω plays the same only one independent variable to describe the role in rotation as the linear velocity v in motion. This in translation corresponds to linear Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 117 motion. This section is limited only to kinematics. We shall turn to dynamics in later sections. We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.6.29) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 6.29 also shows θ0, the angular displacement at t = 0. We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a Fig.6.29 Specifying the angular position of a rigid vector. Further, the angular acceleration, α = body. dω/dt. u Example 6.10 Obtain Eq. (6.36) from first The kinematical quantities in rotational principles. motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) Answer The angular acceleration is uniform, respectively are analogous to kinematic hence quantities in linear motion, displacement (x), dω velocity (v) and acceleration (a). We know the = α = constant (i) kinematical equations of linear motion with d t uniform (i.e. constant) acceleration: Integrating this equation, α dt + c v = v0 + at (a) ω = ∫ 1 2 x = x 0 + υ0t + at (b) = αt + c (as α is constant) 2 At t = 0, ω = ω0 (given) 2 2 υ = υ0 + 2ax (c) From (i) we get at t = 0, ω = c = ω0 Thus, ω = αt + ω0 as required. where x0 = initial displacement and v0= initial With the definition of ω = dθ/dt we may velocity. The word ‘initial’ refers to values of the integrate Eq. (6.36) to get Eq. (6.37). This quantities at t = 0 derivation and the derivation of Eq. (6.38) is left The corresponding kinematic equations for as an exercise. rotational motion with uniform angular acceleration are: u Example 6.11 The angular speed of a motor wheel is increased from 1200 rpm to ω= ω0 + αt (6.36) 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the 1 2 θ = θ0 + ω0t + αt (6.37) acceleration to be uniform? (ii) How many 2 revolutions does the engine make during and ω2 = ω0 2 + 2α(θ– θ0 ) (6.38) this time? Answer where θ0= initial angular displacement of the (i) We shall use ω = ω0 + αt rotating body, and ω0 = initial angular velocity ω0 = initial angular speed in rad/s of the body. Reprint 2025-26 118 PHYSICS = 2π × angular speed in rev/s It is, however, necessary that these correspondences are established on sound 2π × angular speed in rev/min dynamical considerations. This is what we now = 60 s/min turn to. Before we begin, we note a simplification 2π × 1200 that arises in the case of rotational motion = rad/s 60 about a fixed axis. Since the axis is fixed, only those components of torques, which are along = 40π rad/s the direction of the fixed axis need to be Similarly ω = final angular speed in rad/s considered in our discussion. Only these 2π × 3120 components can cause the body to rotate about = rad/s the axis. A component of the torque 60 perpendicular to the axis of rotation will tend to = 2π × 52 rad/s turn the axis from its position. We specifically = 104 π rad/s assume that there will arise necessary forces of constraint to cancel the effect of the ∴Angular acceleration perpendicular components of the (external) torques, so that the fixed position of the axis ω − ω will be maintained. The perpendicular α = 0 = 4 π rad/s2 t components of the torques, therefore need not be taken into account. This means that for our The angular acceleration of the engine calculation of torques on a rigid body: = 4π rad/s2 (1) We need to consider only those forces that (ii) The angular displacement in time t is lie in planes perpendicular to the axis. given by Forces which are parallel to the axis will give torques perpendicular to the axis and need 1 2 θ = ω0 t + αt not be taken into account. 2 (2) We need to consider only those components 1 2 of the position vectors which are = (40π × 16 + × 4π × 16 ) rad 2 perpendicular to the axis. Components of position vectors along the axis will result in = (640π + 512π) rad torques perpendicular to the axis and need = 1152π rad not be taken into account. 1152π = 576 ⊳ Work done by a torqueNumber of revolutions = 2π

3.27 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes? 3.28 The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1? 3.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s–1. Calculate k at 318K and Ea. 3.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Answers to Some Intext Questions 3.1 rav = 6.66 × 10–6 Ms–1 3.2 Rate of reaction = rate of diappearance of A = 0.005 mol litre–1min–1 3.3 Order of the reaction is 2.5 3.4 X ® Y Rate = k[X]2 The rate will increase 9 times 3.5 t = 444 s 3.6 1.925 × 10–4 s–1 3.8 Ea = 52.897 kJ mol–1 3.9 1.471 × 10–19 Chemistry 88 Reprint 2025-26 UnitUnitUnitUnit Unit44 TheThe dd-- andand f-f-Objectives After studying this Unit, you will beable to BlockBlock ElementsElements • learn the positions of the d– and f-block elements in the periodic table; Iron, copper, silver and gold are among the transition elements that • know the electronic configurations have played important roles in the development of human civilisation. of the transition (d-block) and the The inner transition elements such as Th, Pa and U are proving inner transition (f-block) elements; excellent sources of nuclear energy in modern times. • appreciate the relative stability of various oxidation states in terms of electrode potential values; The d-block of the periodic table contains the elements of the groups 3-12 in which the d orbitals are• describe the preparation, progressively filled in each of the four long periods. properties, structures and uses of some important compounds The f-block consists of elements in which 4 f and 5 f such as K2Cr2O7 and KMnO4; orbitals are progressively filled. They are placed in a • understand the general separate panel at the bottom of the periodic table. The characteristics of the d– and names transition metals and inner transition metals f–block elements and the general are often used to refer to the elements of d-and horizontal and group trends in f-blocks respectively. them; There are mainly four series of the transition metals, • describe the properties of the 3d series (Sc to Zn), 4d series (Y to Cd), 5d series (La f-block elements and give a and Hf to Hg) and 6d series which has Ac and elements comparative account of the from Rf to Cn. The two series of the inner transition lanthanoids and actinoids with metals; 4f (Ce to Lu) and 5f (Th to Lr) are known as respect to their electronic lanthanoids and actinoids respectively. configurations, oxidation states Originally the name transition metals was derived and chemical behaviour. from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions. Zinc, cadmium and mercury of group 12 have full d10 configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively, their chemistry is studied along with the chemistry of the transition metals. The presence of partly filled d or f orbitals in their atoms makes transition elements different from that of Reprint 2025-26 the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non- transition elements can be applied successfully to the transition elements also. Various precious metals such as silver, gold and platinum and industrially important metals like iron, copper and titanium belong to the transition metals series. In this Unit, we shall first deal with the electronic configuration, occurrence and general characteristics of transition elements with special emphasis on the trends in the properties of the first row (3d) transition metals along with the preparation and properties of some important compounds. This will be followed by consideration of certain general aspects such as electronic configurations, oxidation states and chemical reactivity of the inner transition metals. THE TRANSITION ELEMENTS (d-BLOCK) 4.14.14.14.14.1 PositionPositionPositionPositionPosition ininininin thethethethethe The d–block occupies the large middle section of the periodic table PeriodicPeriodicPeriodicPeriodicPeriodic TableTableTableTableTable flanked between s– and p– blocks in the periodic table. The d–orbitals of the penultimate energy level of atoms receive electrons giving rise to four rows of the transition metals, i.e., 3d, 4d, 5d and 6d. All these series of transition elements are shown in Table 4.1. 4.24.24.24.24.2 ElectronicElectronicElectronicElectronicElectronic In general1– the electronic configuration of outer orbitals of these elements is (n-1)d 10ns1–2except for Pd where its electronic configuration is 4d105s0. ConfigurationsConfigurationsConfigurationsConfigurationsConfigurations The (n–1) stands for the inner d orbitals which may have one to ten ofofofofof thethethethethe d-Blockd-Blockd-Blockd-Blockd-Block electrons and the outermost ns orbital may have one or two electrons. ElementsElementsElementsElementsElements However, this generalisation has several exceptions because of very little energy difference between (n-1)d and ns orbitals. Furthermore, half and completely filled sets of orbitals are relatively more stable. A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d series. For example, consider the case of Cr, which has 3d 5 4s 1 configuration instead of 3d44s 2; the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals. Similarly in case of Cu, the configuration is 3d104s 1 and not 3d 94s2. The ground state electronic configurations of the outer orbitals of transition elements are given in Table 4.1. Table 4.1: Electronic Configurations of outer orbitals of the Transition Elements (ground state) 1st Series Sc Ti V Cr Mn Fe Co Ni Cu Zn Z 21 22 23 24 25 26 27 28 29 30 4s 2 2 2 1 2 2 2 2 1 2 3d 1 2 3 5 5 6 7 8 10 10 Chemistry 90 Reprint 2025-26 2nd Series Y Zr Nb Mo Tc Ru Rh Pd Ag Cd Z 39 40 41 42 43 44 45 46 47 48 5s 2 2 1 1 1 1 1 0 1 2 4d 1 2 4 5 6 7 8 10 10 10 3rd Series La Hf Ta W Re Os Ir Pt Au Hg Z 57 72 73 74 75 76 77 78 79 80 6s 2 2 2 2 2 2 2 1 1 2 5d 1 2 3 4 5 6 7 9 10 10 4th Series Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Z 89 104 105 106 107 108 109 110 111 112 7s 2 2 2 2 2 2 2 2 1 2 6d 1 2 3 4 5 6 7 8 10 10 The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n-1)d 10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. The d orbitals of the transition elements protrude to the periphery of an atom more than the other orbitals (i.e., s and p), hence, they are more influenced by the surroundings as well as affect the atoms or molecules surrounding them. In some respects, ions of a given dn configuration (n = 1 – 9) have similar magnetic and electronic properties. With partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. All these characteristics have been discussed in detail later in this Unit. There are greater similarities in the properties of the transition elements of a horizontal row in contrast to the non-transition elements. However, some group similarities also exist. We shall first study the general characteristics and their trends in the horizontal rows (particularly 3d row) and then consider some group similarities. On what ground can you say that scandium (Z = 21) is a transition ExampleExampleExampleExampleExample 4.14.14.14.14.1 element but zinc (Z = 30) is not? On the basis of incompletely filled 3d orbitals in case of scandium atom SolutionSolutionSolutionSolutionSolution in its ground state (3d1), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d10) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element. 91 The d- and f- Block Elements Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? We will discuss the properties of elements of first transition series only in the following sections. 4.34.34.34.34.3 GeneralGeneralGeneralGeneralGeneral 4.3.1 Physical Properties PropertiesPropertiesPropertiesPropertiesProperties ofofofofof Nearly all the transition elements display typical metallic properties thethethethethe TransitionTransitionTransitionTransitionTransition such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. With the exceptions of Zn, ElementsElementsElementsElementsElements Cd, Hg and Mn, they have one or more typical metallic structures at (d-Block)(d-Block)(d-Block)(d-Block)(d-Block) normal temperatures. Lattice Structures of Transition Metals Sc Ti V Cr Mn Fe Co Ni Cu Zn hcp hcp bcc bcc X bcc ccp ccp ccp X (bcc) (bcc) (bcc, ccp) (hcp) (hcp) (hcp) Y Zr Nb Mo Tc Ru Rh Pd Ag Cd hcp hcp bcc bcc hcp hcp ccp ccp ccp X (bcc) (bcc) (hcp) La Hf Ta W Re Os Ir Pt Au Hg hcp hcp bcc bcc hcp hcp ccp ccp ccp X (ccp,bcc) (bcc) 4 (bcc = body centred cubic; hcp = hexagonal close packed; ccp = cubic close packed; X = a typical metal structure). W The transition metals (with the exception Re Ta of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are 3 Mo Os high. Fig. 4.1 depicts the melting points of Nb Ru transition metals belonging to 3d, 4d and 5d Ir series. The high melting points of these metals Hf Tc K are attributed to the involvement of greater 3 Cr Rh number of electrons from (n-1)d in addition to Zr V Pt 2 the ns electrons in the interatomic metallic bonding. In any row the melting points of these M.p./10 Ti Fe Co Pd 5 metals rise to a maximum at d except for Ni anomalous values of Mn and Tc and fall Mn Cu regularly as the atomic number increases. Au Ag They have high enthalpies of atomisation which 1 are shown in Fig. 4.2. The maxima at about Atomic number the middle of each series indicate that one Fig. 4.1: Trends in melting points of unpaired electron per d orbital is particularly transition elements Chemistry 92 Reprint 2025-26 favourable for strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Since the enthalpy of atomisation is an important factor in determining the standard electrode potential of a metal, metals with very high enthalpy of atomisation (i.e., very high boiling point) tend to be noble in their reactions (see later for electrode potentials). Another generalisation that may be drawn from Fig. 4.2 is that the metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series; this is an important factor in accounting for the occurrence of much more frequent metal – metal bonding in compounds of the heavy transition metals. –1 mol V/kJ DaH Fig. 4.2 Trends in enthalpies of atomisation of transition elements 4.3.2 Variation in In general, ions of the same charge in a given series show progressive Atomic and decrease in radius with increasing atomic number. This is because the Ionic Sizes new electron enters a d orbital each time the nuclear charge increases of by unity. It may be recalled that the shielding effect of a d electron is Transition not that effective, hence the net electrostatic attraction between the Metals nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. The curves in Fig. 4.3 show an increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected 93 The d- and f- Block Elements Reprint 2025-26 increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship. 19 The factor responsible for the lanthanoid 18 contraction is somewhat similar to that observed in an ordinary transition series and is attributed 17 to similar cause, i.e., the imperfect shielding of 16 one electron by another in the same set of orbitals. However, the shielding of one 4f electron by 15 Radius/nm another is less than that of one d electron by 14 another, and as the nuclear charge increases 13 along the series, there is fairly regular decrease in the size of the entire 4f n orbitals. 12 Sc Ti V Cr Mn Fe Co Ni Cu Zn The decrease in metallic radius coupled with Y Zr Nb Mo Tc Ru Rh Pd Ag Cd increase in atomic mass results in a general increase in the density of these elements. Thus, La Hf Ta W Re Os Ir Pt Au Hg from titanium (Z = 22) to copper (Z = 29) the Fig. 4.3: Trends in atomic radii of significant increase in the density may be noted transition elements (Table 4.2). Table 4.2: Electronic Configurations and some other Properties of the First Series of Transition Elements Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic number 21 22 23 24 25 26 27 28 29 30 Electronic configuration M 3d 14s 2 3d 24s 2 3d 34s 2 3d 54s 1 3d 54s 2 3d 64s 2 3d 74s 2 3d 84s 2 3d 104s 1 3d 104s 2 M + 3d 14s 1 3d 24s 1 3d 34s 1 3d 5 3d 54s 1 3d 64s 1 3d 74s 1 3d 84s 1 3d 10 3d 104s 1 M 2+ 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10 M 3+ [Ar] 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 – – Enthalpy of atomisation, DaH o/kJ mol–1 326 473 515 397 281 416 425 430 339 126 Ionisation enthalpy/DiH o/kJ mol –1 DiHo I 631 656 650 653 717 762 758 736 745 906 DiHo II 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734 DiHo III 2393 2657 2833 2990 3260 2962 3243 3402 3556 3837 Metallic/ionic M 164 147 135 129 137 126 125 125 128 137 radii/pm M 2+ – – 79 82 82 77 74 70 73 75 M 3+ 73 67 64 62 65 65 61 60 – – Standard electrode M 2+/M – –1.63 –1.18 –0.90 –1.18 –0.44 –0.28 –0.25 +0.34 -0.76 potential Eo/V M 3+/M 2+ – –0.37 –0.26 –0.41 +1.57 +0.77 +1.97 – – – Density/g cm –3 3.43 4.1 6.07 7.19 7.21 7.8 8.7 8.9 8.9 7.1 Chemistry 94 Reprint 2025-26 Why do the transition elements exhibit higher enthalpies of ExampleExampleExampleExampleExample 4.24.24.24.24.2 atomisation? Because of large number of unpaired electrons in their atoms they SolutionSolutionSolutionSolutionSolution have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why? 4.3.3 Ionisation There is an increase in ionisation enthalpy along each series of the Enthalpies transition elements from left to right due to an increase in nuclear charge which accompanies the filling of the inner d orbitals. Table 4.2 gives the values of the first three ionisation enthalpies of the first series of transition elements. These values show that the successive enthalpies of these elements do not increase as steeply as in the case of non-transition elements. The variation in ionisation enthalpy along a series of transition elements is much less in comparison to the variation along a period of non-transition elements. The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series. The irregular trend in the first ionisation enthalpy of the metals of 3d series, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals. You have learnt that when d-block elements form ions, ns electrons are lost before (n – 1) d electrons. As we move along the period in 3d series, we see that nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner subshell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly. Thus, ionization energies increase only slightly along the 3d series. The doubly or more highly charged ions have dn configurations with no 4s electrons. A general trend of increasing values of second ionisation enthalpy is expected as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction. However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn2+ and Fe3+ respectively. In both the cases, ions have d5 configuration. Similar breaks occur at corresponding elements in the later transition series. The interpretation of variation in ionisation enthalpy for an electronic configuration dn is as follows: The three terms responsible for the value of ionisation enthalpy are attraction of each electron towards nucleus, repulsion between the 95 The d- and f- Block Elements Reprint 2025-26 electrons and the exchange energy. Exchange energy is responsible for the stabilisation of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hunds rule). The loss of exchange energy increases the stability. As the stability increases, the ionisation becomes more difficult. There is no loss of exchange energy at d6 configuration. Mn+ has 3d54s1 configuration and configuration of Cr+ is d5, therefore, ionisation enthalpy of Mn+ is lower than Cr+. In the same way, Fe2+ has d6 configuration and Mn2+ has 3d5 configuration. Hence, ionisation enthalpy of Fe2+ is lower than the Mn2+. In other words, we can say that the third ionisation enthalpy of Fe is lower than that of Mn. The lowest common oxidation state of these metals is +2. To form the M 2+ ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthalpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M + ions have the d 5 and d 10 configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of one 4s electron which results in the formation of stable d 10 configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d 5 (Mn 2+) and d 10 (Zn 2+) ions. In general, the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Although ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, this problem is very complex and not amenable to ready generalisation. 4.3.4 Oxidation One of the notable features of a transition elements is the great variety States of oxidation states these may show in their compounds. Table 4.3 lists the common oxidation states of the first row transition elements. Table 4.3: Oxidation States of the first row Transition Metal (the most common ones are in bold types) Sc Ti V Cr Mn Fe Co Ni Cu Zn +2 +2 +2 +2 +2 +2 +2 +1 +2 +3 +3 +3 +3 +3 +3 +3 +3 +2 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +6 +7 Chemistry 96 Reprint 2025-26 The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus, early in the series scandium(II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II). At the other end, the only oxidation state of zinc is +2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese (Ti IVO2, VVO2 +, Cr V1O42–, MnVIIO4–) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are FeII,III, Co II,III, NiII, CuI,II, Zn II. The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity, e.g., V II, V III, VIV, VV. This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit of two. An interesting feature in the variability of oxidation states of the d– block elements is noticed among the groups (groups 4 through 10). Although in the p–block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not. Low oxidation states are found when a complex compound has ligands capable of p-acceptor character in addition to the s-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero. Name a transition element which does not exhibit variable ExampleExampleExampleExampleExample 4.34.34.34.34.3 oxidation states. Scandium (Z = 21) does not exhibit variable oxidation states. SolutionSolutionSolutionSolutionSolution IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? 97 The d- and f- Block Elements Reprint 2025-26 4.3.5 Trends in the Table 4.4 contains the thermochemical parameters related to the M2+/M transformation of the solid metal atoms to M2+ ions in solution and their V Standard standard electrode potentials. The observed values of E and those Electrode calculated using the data of Table 4.4 are compared in Fig. 4.4. Potentials The unique behaviour of Cu, having a positive EV, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration V enthalpy. The general trend towards less negative E values across the Fig. 4.4: Observed and calculated values for the standard electrode potentials (M2+ ® M°) of the elements Ti to Zn series is related to the general increase in the sum of the first and second V ionisation enthalpies. It is interesting to note that the value of E for Mn, Ni and Zn are more negative than expected from the trend. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration? ExampleExampleExampleExampleExample 4.44.44.44.44.4 Cr 2+ is reducing as its configuration changes from d 4 to d 3, the latter SolutionSolutionSolutionSolutionSolution having a half-filled t2g level (see Unit 5). On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.4 The E o(M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high DaH o and low DhydH o) Chemistry 98 Reprint 2025-26 Table 4.4: Thermochemical data (kJ mol-1) for the first row Transition Elements and the Standard Electrode Potentials for the Reduction of MII to M. Element (M) DaH o (M) DiH1o D1H2o DhydH o(M2+) Eo/V Ti 469 656 1309 -1866 -1.63 V 515 650 1414 -1895 -1.18 Cr 398 653 1592 -1925 -0.90 Mn 279 717 1509 -1862 -1.18 Fe 418 762 1561 -1998 -0.44 Co 427 758 1644 -2079 -0.28 Ni 431 736 1752 -2121 -0.25 Cu 339 745 1958 -2121 0.34 Zn 130 906 1734 -2059 -0.76 The stability of the half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are related to their E o values, whereas E o for Ni is related to the highest negative DhydH o. 4.3.6 Trends in An examination of the E o(M3+/M2+) values (Table 4.2) shows the varying the M3+/M2+ trends. The low value for Sc reflects the stability of Sc3+ which has a Standard noble gas configuration. The highest value for Zn is due to the removal Electrode of an electron from the stable d 10 configuration of Zn 2+. The Potentials comparatively high value for Mn shows that Mn 2+(d5) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe 3+ (d5). The comparatively low value for V is related to the stability of V 2+ (half-filled t2g level, Unit 5). 4.3.7 Trends in Table 4.5 shows the stable halides of the 3d series of transition metals. Stability of The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 Higher and CrF6. The +7 state for Mn is not represented in simple halides but Oxidation MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 States and CoF3. The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6. Although V +5 is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides is their instability in the low oxidation states e.g., VX2 (X = CI, Br or I) Table 4.5: Formulas of Halides of 3d Metals Oxidation Number + 6 CrF6 + 5 VF5 CrF5 + 4 TiX4 VXI4 CrX4 MnF4 + 3 TiX3 VX3 CrX3 MnF3 FeXI3 CoF3 + 2 TiX2III VX2 CrX2 MnX2 FeX2 CoX2 NiX2 CuX2II ZnX2 + 1 CuXIII Key: X = F ® I; XI = F ® Br; XII = F, CI; XIII = CI ® I 99 The d- and f- Block Elements Reprint 2025-26 and the same applies to CuX. On the other hand, all Cu II halides are known except the iodide. In this case, Cu 2+ oxidises I – to I2: 2Cu 2   4I   Cu2 I2 s  I2 However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation. 2Cu + ® Cu 2+ + Cu The stability of Cu 2+ (aq) rather than Cu+(aq) is due to the much more negative DhydH o of Cu 2+ (aq) than Cu +, which more than compensates for the second ionisation enthalpy of Cu. The ability of oxygen to stabilise the highest oxidation state is demonstrated in the oxides. The highest oxidation number in the oxides (Table 4.6) coincides with the group number and is attained in Sc2O3 to Mn2O7. Beyond Group 7, no higher oxides of Fe above Fe2O3, are known, although ferrates (VI)(FeO4)2–, are formed in alkaline media but they readily decompose to Fe2O3 and O2. Besides the oxides, oxocations stabilise V v as VO2 +, V IV as VO2+ and Ti IV as TiO 2+. The ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. Thus the highest Mn fluoride is MnF4 whereas the highest oxide is Mn2O7. The ability of oxygen to form multiple bonds to metals explains its superiority. In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O’s including a Mn–O–Mn bridge. The tetrahedral [MO4]n- ions are known for V V, Cr Vl, Mn V, Mn Vl and Mn VII. Table 4.6: Oxides of 3d Metals Oxidation Groups Number 3 4 5 6 7 8 9 10 11 12 + 7 Mn2O7 + 6 CrO3 + 5 V2O5 + 4 TiO2 V2O4 CrO2 MnO2 + 3 Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Mn3O4* Fe3O4 * Co3O4* + 2 TiO VO (CrO) MnO FeO CoO NiO CuO ZnO + 1 Cu2O * mixed oxides How would you account for the increasing oxidising power in the ExampleExampleExampleExampleExample 4.54.54.54.54.5 series VO2+ < Cr2O7 2– < MnO4 – ? This is due to the increasing stability of the lower species to which they SolutionSolutionSolutionSolutionSolution are reduced. IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements? Chemistry 100 Reprint 2025-26 4.3.8 Chemical Transition metals vary widely in their chemical reactivity. Many of Reactivity them are sufficiently electropositive to dissolve in mineral acids, although and Eo a few are ‘noble’—that is, they are unaffected by single acids. Values The metals of the first series with the exception of copper are relatively more reactive and are oxidised by 1M H +, though the actual rate at which these metals react with oxidising agents like hydrogen ion (H +) is sometimes slow. For example, titanium and vanadium, in practice, are passive to dilute non oxidising acids at room temperature. The E o values for M2+/M (Table 4.2) indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative E o values is related to the increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the E o values for Mn, Ni and Zn are more negative than expected from the general trend. Whereas the stabilities of half-filled d subshell (d5) in Mn2+ and completely filled d subshell (d10) in zinc are related to their E e values; for nickel, Eo value is related to the highest negative enthalpy of hydration. An examination of the E o values for the redox couple M 3+/M2+ (Table 4.2) shows that Mn 3+ and Co 3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti 2+, V 2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g., 2 Cr 2+(aq) + 2 H+(aq) ® 2 Cr 3+(aq) + H2(g) ExampleExampleExampleExampleExample 4.64.64.64.64.6 For the first row transition metals the Eo values are: E o V Cr Mn Fe Co Ni Cu (M2+/M) –1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34 Explain the irregularity in the above values. SolutionSolutionSolutionSolutionSolution The E o (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (  i H1  i H 2 ) and also the sublimation enthalpies which are relatively much less for manganese and vanadium. ExampleExampleExampleExampleExample 4.74.74.74.74.7 Why is the E o value for the Mn3+/Mn 2+ couple much more positive than that for Cr 3+/Cr2+ or Fe 3+/Fe 2+? Explain. SolutionSolutionSolutionSolutionSolution Much larger third ionisation energy of Mn (where the required change is d5 to d4) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? 4.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ? 4.3.9 Magnetic When a magnetic field is applied to substances, mainly two types of Properties magnetic behaviour are observed: diamagnetism and paramagnetism. Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. Substances which are 101 The d- and f- Block Elements Reprint 2025-26 attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic. Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e.,  n  n  2  where n is the number of unpaired electrons and µ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM). The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion. The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 4.7. The experimental data are mainly for hydrated ions in solution or in the solid state. Table 4.7: Calculated and Observed Magnetic Moments (BM) Ion Configuration Unpaired Magnetic moment electron(s) Calculated Observed Sc3+ 3d0 0 0 0 Ti 3+ 3d1 1 1.73 1.75 Tl2+ 3d2 2 2.84 2.76 V2+ 3d3 3 3.87 3.86 Cr2+ 3d4 4 4.90 4.80 Mn2+ 3d5 5 5.92 5.96 Fe2+ 3d6 4 4.90 5.3 – 5.5 Co2+ 3d7 3 3.87 4.4 – 5.2 Ni2+ 3d8 2 2.84 2.9 – 3, 4 Cu 2+ 3d9 1 1.73 1.8 – 2.2 Zn2+ 3d10 0 0 Calculate the magnetic moment of a divalent ion in aqueous solution ExampleExampleExampleExampleExample 4.84.84.84.84.8 if its atomic number is 25. With atomic number 25, the divalent ion in aqueous solution will have SolutionSolutionSolutionSolutionSolution d5 configuration (five unpaired electrons). The magnetic moment, µ is  5  5  2   5.92BM Chemistry 102 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.8 Calculate the ‘spin only’ magnetic moment of M 2+ (aq) ion (Z = 27). 4.3.10 Formation When an electron from a lower energy d orbital is excited to a higher of Coloured energy d orbital, the energy of excitation corresponds to the frequency Ions of light absorbed (Unit 5). This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. In aqueous solutions where water molecules are the ligands, the colours of the ions observed are listed in Table 4.8. A few coloured solutions of Fig. 4.5: Colours of some of the first row d–block elements are transition metal ions in aqueous solutions. From illustrated in Fig. 4.5. left to right: V4+,V3+,Mn2+,Fe3+,Co2+,Ni2+and Cu2+ . Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions Configuration Example Colour 3d0 Sc3+ colourless 3d0 Ti 4+ colourless 3d1 Ti 3+ purple 3d1 V4+ blue 3d2 V3+ green 3d3 V2+ violet 3d3 Cr3+ violet 3d4 Mn 3+ violet 3d4 Cr2+ blue 3d5 Mn 2+ pink 3d5 Fe3+ yellow 3d6 Fe2+ green 3d63d7 Co3+Co2+ bluepink 3d8 Ni2+ green 3d9 Cu 2+ blue 3d10 Zn2+ colourless 4.3.11 Formation Complex compounds are those in which the metal ions bind a number of Complex of anions or neutral molecules giving complex species with Compounds characteristic properties. A few examples are: [Fe(CN)6] 3–, [Fe(CN)6]4–, [Cu(NH3)4] 2+ and [PtCl4] 2–. (The chemistry of complex compounds is 103 The d- and f- Block Elements Reprint 2025-26 dealt with in detail in Unit 5). The transition metals form a large number of complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d orbitals for bond formation. 4.3.12 Catalytic The transition metals and their compounds are known for their catalytic Properties activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in Catalytic Hydrogenation) are some of the examples. Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions. 2 I– + S2O8 2– ® I2 + 2 SO4 2– An explanation of this catalytic action can be given as: 2 Fe 3+ + 2 I – ® 2 Fe 2+ + I2 2 Fe 2+ + S2O82– ® 2 Fe3+ + 2SO42– 4.3.13 Formation Interstitial compounds are those which are formed when small atoms of like H, C or N are trapped inside the crystal lattices of metals. They are Interstitial usually non stoichiometric and are neither typically ionic nor covalent, Compounds for example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. The formulas quoted do not, of course, correspond to any normal oxidation state of the metal. Because of the nature of their composition, these compounds are referred to as interstitial compounds. The principal physical and chemical characteristics of these compounds are as follows: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard, some borides approach diamond in hardness. (iii) They retain metallic conductivity. (iv) They are chemically inert. 4.3.14 Alloy An alloy is a blend of metals prepared by mixing the components. Formation Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals. The alloys so formed are hard and have often high melting points. The best known are ferrous alloys: chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel. Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance. Chemistry 104 Reprint 2025-26 ExampleExampleExampleExampleExample 4.94.94.94.94.9 What is meant by ‘disproportionation’ of an oxidation state? Give an example. SolutionSolutionSolutionSolutionSolution When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution. 3 Mn VIO4 2– + 4 H + ® 2 Mn VIIO –4 + Mn IVO2 + 2H2O IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.9 Explain why Cu+ ion is not stable in aqueous solutions? 4.44.44.44.44.4 SomeSomeSomeSomeSome 4.4.1 Oxides and Oxoanions of Metals ImportantImportantImportantImportantImportant These oxides are generally formed by the reaction of metals with CompoundsCompoundsCompoundsCompoundsCompounds ofofofofof oxygen at high temperatures. All the metals except scandium form TransitionTransitionTransitionTransitionTransition MO oxides which are ionic. The highest oxidation number in the oxides, coincides with the group number and is attained in Sc2O3 to ElementsElementsElementsElementsElements Mn2O7. Beyond group 7, no higher oxides of iron above Fe2O3 are known. Besides the oxides, the oxocations stabilise V V as VO2 +, V IV as VO 2+ and Ti IV as TiO 2+. As the oxidation number of a metal increases, ionic character decreases. In the case of Mn, Mn2O7 is a covalent green oil. Even CrO3 and V2O5 have low melting points. In these higher oxides, the acidic character is predominant. Thus, Mn2O7 gives HMnO4 and CrO3 gives H2CrO4 and H2Cr2O7. V2O5 is, however, amphoteric though mainly acidic and it gives VO4 3– as well as VO2+ salts. In vanadium there is gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO 2+ salts. Similarly, V2O5 reacts with alkalies as well as acids to give VO 34  and VO4 respectively. The well characterised CrO is basic but Cr2O3 is amphoteric. Potassium dichromate K2Cr2O7 Potassium dichromate is a very important chemical used in leather industry and as an oxidant for preparation of many azo compounds. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows: 4 FeCr2O4 + 8 Na2CO3 + 7 O2 ® 8 Na2CrO4 + 2 Fe2O3 + 8 CO2 The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised. 2Na2CrO4 + 2 H+ ® Na2Cr2O7 + 2 Na + + H2O 105 The d- and f- Block Elements Reprint 2025-26 Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2 KCl ® K2Cr2O7 + 2 NaCl Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same. 2 CrO4 2– + 2H + ® Cr2O7 2– + H2O Cr2O7 2– + 2 OH- ® 2 CrO4 2– + H2O The structures of chromate ion, CrO4 2– and the dichromate ion, Cr2O7 2– are shown below. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr–O–Cr bond angle of 126°. Sodium and potassium dichromates are strong oxidising agents; the sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis. In acidic solution, its oxidising action can be represented as follows: Cr2O7 2– + 14H + + 6e – ® 2Cr 3+ + 7H2O (E o = 1.33V) Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III). The half-reactions are noted below: 6 I– ® 3I2 + 6 e – ; 3 Sn 2+ ® 3Sn 4+ + 6 e – 3 H2S ® 6H+ + 3S + 6e – ; 6 Fe 2+ ® 6Fe3+ + 6 e– The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g., Cr2O7 2– + 14 H+ + 6 Fe2+ ® 2 Cr3+ + 6 Fe3+ + 7 H2O Potassium permanganate KMnO4 Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. 2MnO2 + 4KOH + O2 ® 2K2MnO4 + 2H2O 3MnO4 2– + 4H+ ® 2MnO4 – + MnO2 + 2H2O Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (Vl). F used with KOH, oxidised Electrolytic oxidation in MnO 2 →with air or KNO 3 MnO 24 − ; MnO 24  alkaline solution MnO 4 manganate ion manganate permanganate ion Chemistry 106 Reprint 2025-26 In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate. 2Mn2+ + 5S2O8 2– + 8H2O ® 2MnO4 – + 10SO42– + 16H + Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of KClO4. The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K. 2KMnO4 ® K2MnO4 + MnO2 + O2 It has two physical properties of considerable interest: its intense colour and its diamagnetism along with temperature-dependent weak paramagnetism. These can be explained by the use of molecular orbital theory which is beyond the present scope. The manganate and permanganate ions are tetrahedral; the p- bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese. The green manganate is paramagnetic because of one unpaired electron but the permanganate is diamagnetic due to the absence of unpaired electron. Acidified permanganate solution oxidises oxalates to carbon dioxide, iron(II) to iron(III), nitrites to nitrates and iodides to free iodine. The half-reactions of reductants are: COO – 5 10CO2 + 10e – COO – 5 Fe2+ ® 5 Fe3+ + 5e– 5NO2 – + 5H2O ® 5NO3 – + 10H+ + l0e– 10I– ® 5I2 + 10e– The full reaction can be written by adding the half-reaction for KMnO4 to the half-reaction of the reducing agent, balancing wherever necessary. If we represent the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt by half-reactions, MnO4 – + e– ® MnO4 2– (E o = + 0.56 V) MnO4 – + 4H+ + 3e– ® MnO2 + 2H2O (E o = + 1.69 V) MnO4 – + 8H+ + 5e– ® Mn2+ + 4H2O (E o = + 1.52 V) We can very well see that the hydrogen ion concentration of the solution plays an important part in influencing the reaction. Although many reactions can be understood by consideration of redox potential, kinetics of the reaction is also an important factor. Permanganate at [H+] = 1 should oxidise water but in practice the reaction is extremely slow unless either manganese(ll) ions are present or the temperature is raised. A few important oxidising reactions of KMnO4 are given below: 1. In acid solutions: (a) Iodine is liberated from potassium iodide : 10I – + 2MnO4 – + 16H + ® 2Mn2+ + 8H2O + 5I2 (b) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe 2+ + MnO4 – + 8H+ ® Mn2+ + 4H2O + 5Fe 3+ 107 The d- and f- Block Elements Reprint 2025-26 (c) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O4 2– + 2MnO4 – + 16H + ——> 2Mn 2+ + 8H2O + 10CO2 (d) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H + + S2– 5S 2– + 2MnO – 4 + 16H + ——> 2Mn2+ + 8H2O + 5S (e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid: 5SO3 2– + 2MnO4 – + 6H + ——> 2Mn 2+ + 3H2O + 5SO42– (f) Nitrite is oxidised to nitrate: 5NO2– + 2MnO4– + 6H + ——> 2Mn 2+ + 5NO3 – + 3H2O 2. In neutral or faintly alkaline solutions: (a) A notable reaction is the oxidation of iodide to iodate: 2MnO4 – + H2O + I– ——> 2MnO2 + 2OH – + IO3 – (b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4 – + 3S2O3 2– + H2O ——> 8MnO2 + 6SO4 2– + 2OH – (c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation: 2MnO4 – + 3Mn 2+ + 2H2O ——> 5MnO2 + 4H+ Note: Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine. UsesUsesUses:UsesUses Besides its use in analytical chemistry, potassium permanganate is used as a favourite oxidant in preparative organic chemistry. Its uses for the bleaching of wool, cotton, silk and other textile fibres and for the decolourisation of oils are also dependent on its strong oxidising power. THE INNER TRANSITION ELEMENTS ( f-BLOCK) The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium). Because lanthanum closely resembles the lanthanoids, it is usually included in any discussion of the lanthanoids for which the general symbol Ln is often used. Similarly, a discussion of the actinoids includes actinium besides the fourteen elements constituting the series. The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series. They have only one stable oxidation state and their chemistry provides an excellent opportunity to examine the effect of small changes in size and nuclear charge along a series of otherwise similar elements. The chemistry of the actinoids is, on the other hand, much more complicated. The complication arises partly owing to the occurrence of a wide range of oxidation states in these elements and partly because their radioactivity creates special problems in their study; the two series will be considered separately here. 4.54.54.54.54.5 TheTheTheTheThe The names, symbols, electronic configurations of atomic and some LanthanoidsLanthanoidsLanthanoidsLanthanoidsLanthanoids ionic states and atomic and ionic radii of lanthanum and lanthanoids (for which the general symbol Ln is used) are given in Table 4.9. Chemistry 108 Reprint 2025-26 4.5.1 Electronic It may be noted that atoms of these elements have electronic Configurations configuration with 6s 2 common but with variable occupancy of 4f level (Table 4.9). However, the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form 4f n (n = 1 to 14 with increasing atomic number). 4.5.2 Atomic and The overall decrease in atomic and ionic radii from lanthanum to Ionic Sizes lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. It has far reaching Sm 2+ consequences in the chemistry of the third 110 2+ transition series of the elements. The decrease Eu in atomic radii (derived from the structures of La3+ metals) is not quite regular as it is regular in 3+ M3+ ions (Fig. 4.6). This contraction is, of Ce course, similar to that observed in an ordinary Pr3+ transition series and is attributed to the same 100 Nd3+ cause, the imperfect shielding of one electron Pm 3+ by another in the same sub-shell. However, the Sm3+ shielding of one 4 f electron by another is less Eu3+ than one d electron by another with the increase Gd3+ Tm 2+radii/pm 2+ in nuclear charge along the series. There is Yb Ce 4+ Tb 3+ fairly regular decrease in the sizes with 3+ DyIonic Pr4+ 3+ increasing atomic number. 90 Ho Er 3+ The cumulative effect of the contraction of Tm3+ the lanthanoid series, known as lanthanoid Yb3+ 3+ contraction, causes the radii of the members 4+ Lu Tb of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr 57 59 61 63 65 67 69 71 (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their Atomic number occurrence together in nature and for the Fig. 4.6: Trends in ionic radii of lanthanoids difficulty faced in their separation. 4.5.3 Oxidation In the lanthanoids, La(II) and Ln(III) compounds are predominant States species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained. This irregularity (as in ionisation enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Thus, the formation of Ce IV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. The E o value for Ce 4+/ Ce 3+ is + 1.74 V which suggests that it can oxidise water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent. Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2. Eu2+ is formed by losing the two s electrons and its f 7 configuration accounts for the formation of this ion. However, Eu 2+ is a strong reducing agent changing to the common +3 state. Similarly Yb 2+ which has f 14 configuration is a reductant. Tb IV has half-filled f-orbitals and is an oxidant. The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states. 109 The d- and f- Block Elements Reprint 2025-26 Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids Electronic configurations* Radii/pm Atomic Name Symbol Ln Ln2+ Ln3+ Ln4+ Ln Ln3+ Number 57 Lanthanum La 5d16s2 5d1 4f 0 187 106 58 Cerium Ce 4f15d16s2 4f 2 4f 1 4f 0 183 103 59 Praseodymium Pr 4f 36s2 4f 3 4f 2 4f 1 182 101 60 Neodymium Nd 4f 46s2 4f 4 4f 3 4f 2 181 99 61 Promethium Pm 4f 56s2 4f 5 4f 4 181 98 62 Samarium Sm 4f 66s2 4f 6 4f 5 180 96 63 Europium Eu 4f 76s2 4f 7 4f 6 199 95 64 Gadolinium Gd 4f 75d16s2 4f 75d 1 4f 7 180 94 65 Terbium Tb 4f 96s2 4f 9 4f 8 4f 7 178 92 66 Dysprosium Dy 4f 106s2 4f 10 4f 9 4f 8 177 91 67 Holmium Ho 4f 116s2 4f 11 4f 10 176 89 68 Erbium Er 4f 126s2 4f 12 4f 11 175 88 69 Thulium Tm 4f 136s2 4f 13 4f 12 174 87 70 Ytterbium Yb 4f 146s2 4f 14 4f 13 173 86 71 Lutetium Lu 4f 145d16s2 4f 145d1 4f 14 – – – * Only electrons outside [Xe] core are indicated 4.5.4 General All the lanthanoids are silvery white soft metals and tarnish rapidly in air. Characteristics The hardness increases with increasing atomic number, samarium being steel hard. Their melting points range between 1000 to 1200 K but samarium melts at 1623 K. They have typical metallic structure and are good conductors of heat and electricity. Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm. Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of f electrons. Neither La 3+ nor Lu3+ ion shows any colour but the rest do so. However, absorption bands are narrow, probably because of the excitation within f level. The lanthanoid ions other than the f 0 type (La 3+ and Ce4+) and the f 14 type (Yb2+ and Lu3+) are all paramagnetic. The first ionisation enthalpies of the lanthanoids are around 600 kJ mol –1, the second about 1200 kJ mol–1 comparable with those of calcium. A detailed discussion of the variation of the third ionisation enthalpies indicates that the exchange enthalpy considerations (as in 3d orbitals of the first transition series), appear to impart a certain degree of stability to empty, half-filled and completely filled orbitals f level. This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium. In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium. Values for E o for the half-reaction: Ln 3+(aq) + 3e – ® Ln(s) Chemistry 110 Reprint 2025-26 Ln2 O 3 H2 are in the range of –2.2 to –2.4 V except for Eu for which the value is – 2.0 V. This is, of course, a small acids variation. The metals combine with burns in with hydrogen when gently heated in the O2 gas. The carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated heated with S Ln with halogens with carbon. They liberate hydrogen Ln 2 S3 LnX 3 from dilute acids and burn in halogens N with with toandform hydroxideshalides. They formM(OH)3.oxides M2O3The C K H2 O hydroxides are definite compounds, not heated just hydrated oxides. They are basic with 2773 like alkaline earth metal oxides and Ln N LnC2 Ln(OH)3 + H2 hydroxides. Their general reactions are depicted in Fig. 4.7. Fig 4.7: Chemical reactions of the lanthanoids. The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking. Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces. 4.64.64.64.64.6 TheTheTheTheThe ActinoidsActinoidsActinoidsActinoidsActinoids The actinoids include the fourteen elements from Th to Lr. The names, symbols and some properties of these elements are given in Table 4.10. Table 4.10: Some Properties of Actinium and Actinoids Electronic conifigurations* Radii/pm Atomic Name Symbol M M3+ M4+ M3+ M4+ Number 89 Actinium Ac 6d 17s 2 5f 0 111 90 Thorium Th 6d 27s 2 5f 1 5f 0 99 91 Protactinium Pa 5f 26d 17s 2 5f 2 5f 1 96 92 Uranium U 5f 36d 17s 2 5f 3 5f 2 103 93 93 Neptunium Np 5f 46d 17s 2 5f 4 5f 3 101 92 94 Plutonium Pu 5f 67s 2 5f 5 5f 4 100 90 95 Americium Am 5f 77s 2 5f 6 5f 5 99 89 96 Curium Cm 5f 76d 17s 2 5f 7 5f 6 99 88 97 Berkelium Bk 5f 97s 2 5f 8 5f 7 98 87 98 Californium Cf 5f 107s 2 5f 9 5f 8 98 86 99 Einstenium Es 5f 117s 2 5f 10 5f 9 – – 100 Fermium Fm 5f 127s 2 5f 11 5f 10 – – 101 Mendelevium Md 5f 137s 2 5f 12 5f 11 – – 102 Nobelium No 5f 147s 2 5f 13 5f 12 – – 103 Lawrencium Lr 5f 146d 17s 2 5f 14 5f 13 – – 111 The d- and f- Block Elements Reprint 2025-26 The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z =103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult. 4.6.1 Electronic All the actinoids are believed to have the electronic configuration of 7s2 Configurations and variable occupancy of the 5f and 6d subshells. The fourteen electrons are formally added to 5f, though not in thorium (Z = 90) but from Pa onwards the 5f orbitals are complete at element 103. The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the f 0, f 7 and f 14 occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 76d17s2. Although the 5f orbitals resemble the 4f orbitals in their angular part of the wave-function, they are not as buried as 4f orbitals and hence 5f electrons can participate in bonding to a far greater extent. 4.6.2 Ionic Sizes The general trend in lanthanoids is observable in the actinoids as well. There is a gradual decrease in the size of atoms or M3+ ions across the series. This may be referred to as the actinoid contraction (like lanthanoid contraction). The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons. 4.6.3 Oxidation There is a greater range of oxidation states, which is in part attributed to States the fact that the 5f, 6d and 7s levels are of comparable energies. The known oxidation states of actinoids are listed in Table 4.11. The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements (Table 4.11). The actinoids resemble the lanthanoids in having more compounds in +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse. Because the distribution of oxidation states among the actinoids is so uneven and so different for the former and later elements, it is unsatisfactory to review their chemistry in terms of oxidation states. Table 4.11: Oxidation States of Actinium and Actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 4.6.4 General The actinoid metals are all silvery in appearance but display Characteristics a variety of structures. The structural variability is obtained and Comparison due to irregularities in metallic radii which are far greater with Lanthanoids than in lanthanoids. Chemistry 112 Reprint 2025-26 The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives a mixture of oxide and hydride and combination with most non metals takes place at moderate temperatures. Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no action. The magnetic properties of the actinoids are more complex than those of the lanthanoids. Although the variation in the magnetic susceptibility of the actinoids with the number of unpaired 5 f electrons is roughly parallel to the corresponding results for the lanthanoids, the latter have higher values. It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids. A comparison of the actinoids with the lanthanoids, with respect to different characteristics as discussed above, reveals that behaviour similar to that of the lanthanoids is not evident until the second half of the actinoid series. However, even the early actinoids resemble the lanthanoids in showing close similarities with each other and in gradual variation in properties which do not entail change in oxidation state. The lanthanoid and actinoid contractions, have extended effects on the sizes, and therefore, the properties of the elements succeeding them in their respective periods. The lanthanoid contraction is more important because the chemistry of elements succeeding the actinoids are much less known at the present time. ExampleExampleExampleExampleExample 4.104.104.104.104.10 Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. SolutionSolutionSolutionSolutionSolution Cerium (Z = 58) IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why? 4.74.74.74.74.7 SomeSomeSomeSomeSome Iron and steels are the most important construction materials. Their ApplicationsApplicationsApplicationsApplicationsApplications production is based on the reduction of iron oxides, the removal of impurities and the addition of carbon and alloying metals such as Cr, Mn ofofofofof d-d-d-d-d- andandandandand and Ni. Some compounds are manufactured for special purposes such as f-Blockf-Blockf-Blockf-Blockf-Block TiO for the pigment industry and MnO2 for use in dry battery cells. The ElementsElementsElementsElementsElements battery industry also requires Zn and Ni/Cd. The elements of Group 11 are still worthy of being called the coinage metals, although Ag and Au 113 The d- and f- Block Elements Reprint 2025-26 are restricted to collection items and the contemporary UK ‘copper’ coins are copper-coated steel. The ‘silver’ UK coins are a Cu/Ni alloy. Many of the metals and/or their compounds are essential catalysts in the chemical industry. V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid. TiCl4 with A1(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene). Iron catalysts are used in the Haber process for the production of ammonia from N2/H2 mixtures. Nickel catalysts enable the hydrogenation of fats to proceed. In the Wacker process the oxidation of ethyne to ethanal is catalysed by PdCl2. Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene. The photographic industry relies on the special light-sensitive properties of AgBr. SummarySummarySummarySummarySummary The d-block consisting of Groups 3-12 occupies the large middle section of the periodic table. In these elements the inner d orbitals are progressively filled. The f-block is placed outside at the bottom of the periodic table and in the elements of this block, 4f and 5f orbitals are progressively filled. Corresponding to the filling of 3d, 4d and 5d orbitals, three series of transition elements are well recognised. All the transition elements exhibit typical metallic properties such as –high tensile strength, ductility, malleability, thermal and electrical conductivity and metallic character. Their melting and boiling points are high which are attributed to the involvement of (n –1) d electrons resulting into strong interatomic bonding. In many of these properties, the maxima occur at about the middle of each series which indicates that one unpaired electron per d orbital is particularly a favourable configuration for strong interatomic interaction. Successive ionisation enthalpies do not increase as steeply as in the main group elements with increasing atomic number. Hence, the loss of variable number of electrons from (n –1) d orbitals is not energetically unfavourable. The involvement of (n–1) d electrons in the behaviour of transition elements impart certain distinct characteristics to these elements. Thus, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties and tendency for the formation of coloured ions, interstitial compounds and complexes. The transition elements vary widely in their chemical behaviour. Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are ‘noble’. Of the first series, with the exception of copper, all the metals are relatively reactive. The transition metals react with a number of non-metals like oxygen, nitrogen, sulphur and halogens to form binary compounds. The first series transition metal oxides are generally formed from the reaction of metals with oxygen at high temperatures. These oxides dissolve in acids and bases to form oxometallic salts. Potassium dichromate and potassium permanganate are common examples. Potassium dichromate is prepared from the chromite ore by fusion with alkali in presence of air and acidifying the extract. Pyrolusite ore (MnO2) is used for the preparation of potassium permanganate. Both the dichromate and the permanganate ions are strong oxidising agents. The two series of inner transition elements, lanthanoids and actinoids constitute the f-block of the periodic table. With the successive filling of the inner orbitals, 4f, there is a gradual decrease in the atomic and ionic sizes of these metals along the series (lanthanoid contraction). This has far reaching consequences in the chemistry of the elements succeeding them. Lanthanum and all the lanthanoids are rather soft white metals. They react easily with water to give solutions giving +3 ions. The principal oxidation state is +3, although +4 and +2 oxidation states are also exhibited by some Chemistry 114 Reprint 2025-26 occasionally. The chemistry of the actinoids is more complex in view of their ability to exist in different oxidation states. Furthermore, many of the actinoid elements are radioactive which make the study of these elements rather difficult. There are many useful applications of the d- and f-block elements and their compounds, notable among them being in varieties of steels, catalysts, complexes, organic syntheses, etc. Exercises 4.1 Write down the electronic configuration of: (i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+ (ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+ 4.2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? 4.3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? 4.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples. 4.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d 3, 3d 5, 3d 8 and 3d 4? 4.6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. 4.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction? 4.8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements? 4.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).