Q30 / 90 · JEE Main 2016 (09 Apr Online) · Chemistry

Identify the correct trend given below:

(Atomic No. = Ti : 22, Cr : 24 and Mo : 42)

A
$∆ o f {[C r {(H_{2} O)}_{6}]}^{2 +} > {[M o {(H_{2} O)}_{6}]}^{2 +}$ and $∆ o f {[T i {(H_{2} O)}_{6}]}^{3 +} > {[T i {(H_{2} O)}_{6}]}^{2 +}$
B
$∆ o f {[C r {(H_{2} O)}_{6}]}^{2 +} > {[M o {(H_{2} O)}_{6}]}^{2 +}$ and $∆ o f {[T i {(H_{2} O)}_{6}]}^{3 +} < {[T i {(H_{2} O)}_{6}]}^{2 +}$
Correct answer
C
$∆ o f {[C r {(H_{2} O)}_{6}]}^{2 +} < {[M o {(H_{2} O)}_{6}]}^{2 +}$ and $∆ o f {[T i {(H_{2} O)}_{6}]}^{3 +} > {[T i {(H_{2} O)}_{6}]}^{2 +}$
D
$∆ o f {[C r {(H_{2} O)}_{6}]}^{2 +} < {[M o {(H_{2} O)}_{6}]}^{2 +}$ and $∆ o f {[T i {(H_{2} O)}_{6}]}^{3 +} < {[T i {(H_{2} O)}_{6}]}^{2 +}$

Solution

As the oxidation state of metal increases the CFSE value increases hence

Δ of {[Ti {(H_{2} O)}_{6}]}^{3 +} > {[Ti {(H_{2} O)}_{6}]}^{2 +}

Δ

increase from 3d to 4d Series. So,

$Δ \propto$ CFSE (Crystal field stabilisation energy)

${Δ of [Cr {(H_{2} O)}_{6}]}^{2 +} < Δ of {[Mo {(H_{2} O)}_{6}]}^{2 +}$

Because here $Δ$ depends on Z_eff & Z_eff of 4d series is more than 3d series

Hence, the correct answer is option: 'C'.