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In the following question match the items given in Columns I and II.

Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code

Column I (Complex ion)

Column II (Hybridisation, number of unpaired electrons)

A. $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$

1. $dsp^{2},1$

B. $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$

2. $sp^{3}d^{2},5$

C. $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$

3. $d^{2}sp^{3},3$

D. $\left [ MnF_{6} \right ]^{4-}$

4. $sp^{3},4$

5. $sp^{3}d^{2},2$

(i) A (3) B (1) C (5) D (2)

(ii) A (4) B (3) C (2) D (1)

(iii) A (3) B (2) C (4) D (1)

(iv) A (4) B (1) C (2) D (3)

Answers (1)

Option (i) is the correct answer.

Explanation : (i) Strong field ligand forms inner orbital complex with hybridisation $d^{2}sp^{3}$

(ii) Weak field ligand forms outer orbital complex with hybridisation $d^{2}sp^{3}$ .

According to VBT, hybridisation and number of unpaired electrons of coordination compounds can be calculated as

(a) $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$

MOEC (Molecular orbital electronic configuration) of $Cr^{3+}$ in $\left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+}$ is

Hybridisation = $d^{2}sp^{3}$

n (number of unpaired electrons) = 3

(b) $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$ is

MOEC of $Co^{2+}$ in $\left [ Co\left ( CN \right )_{4} \right ]^{2-}$ is

Hybridisation = $dsp^{2}$

n (number of unpaired electrons) = 1

(c) $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$

MOEC of $Ni^{2+}$ in $\left [ Ni\left ( NH_{3} \right )_{6} \right ]^{2+}$ is

Hybridisation = $sp^{3}d^{2}$

n (number of unpaired electrons) = 2

(d) $\left [ MnF_{6} \right ]^{4-}$

MOEC of $Mn^{2+}$ in $\left [ MnF_{6} \right ]^{4-}$ is

Hybridisation = $sp^{3}d^{2}$

n (number of unpaired electrons) = 5

Hence , correct choice can be represented by (a).

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