The stepwise formation of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is given below: \(\mathrm{Cu}^{2+}+\mathrm{NH}_{3} \xlongequal{\mathrm{~K}_{1}}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)\right]^{2+}\) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)\right]^{2+}+\mathrm{NH}_{3} \xlongequal{\mathrm{~K}_{2}}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{2+}\) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}\right]^{2+}+\mathrm{NH}_{3} \xlongequal{\mathrm{~K}_{3}}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}\right]^{2+}\) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}\right]^{2+}+\mathrm{NH}_{3} \xlongequal{\mathrm{~K}_{4}}\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) The value of stability constants \(K_{1}, K_{2}, K_{3}\) and \(K_{4}\) are \(10^{4}, 1.58 \times 10^{3}, 5 \times 10^{2}\) and \(10^{2}\) respectively. The overall equilibrium constant for dissociation of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is \(x \times 10^{-12}\). The value of \(x\) is ____. (Rounded off to the nearest integer)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Coordination Compounds

The stepwise formation of

[Cu (NH_{3})_{4}]^{2 +}

is given below:

Cu^{2 +} + NH_{3} K_{1} [Cu (NH_{3})]^{2 +}

[Cu (NH_{3})]^{2 +} + NH_{3} K_{2} [Cu (NH_{3})_{2}]^{2 +}

[Cu (NH_{3})_{2}]^{2 +} + NH_{3} K_{3} [Cu (NH_{3})_{3}]^{2 +}

[Cu (NH_{3})_{3}]^{2 +} + NH_{3} K_{4} [Cu (NH_{3})_{4}]^{2 +}

The value of stability constants

K_{1}, K_{2}, K_{3}

and

K_{4}

are

1 0^{4}, 1.58 \times 1 0^{3}, 5 \times 1 0^{2}

and

1 0^{2}

respectively. The overall equilibrium constant for dissociation of

[Cu (NH_{3})_{4}]^{2 +}

x \times 1 0^{- 12}

. The value of

x

is ____. (Rounded off to the nearest integer)

Answer: 1

Solution

The overall dissociation of

[Cu (N H_{3})_{4}]^{2 +}

into

C u^{2 +}

and four

N H_{3}

is the reverse of the overall formation, so its equilibrium constant is the reciprocal of the overall stability constant. The overall stability constant is the product of the four stepwise constants,

K_{f} = K_{1} K_{2} K_{3} K_{4} = 1 0^{4} \times (1.58 \times 1 0^{3}) \times (5 \times 1 0^{2}) \times 1 0^{2} = 7.9 \times 1 0^{11}

. The dissociation constant is therefore

K_{d} = \frac{1}{K _{f}} = \frac{1}{7.9 \times 1 0 ^{11}} = 1.27 \times 1 0^{- 12}

. Written in the form

x \times 1 0^{- 12}

, this gives

x = 1.27

, which rounds to

1

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Coordination Compounds →

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Coordination Compounds

Solution

More JEE Main questions on Coordination Compounds