What is(are) number(s) of unpaired electrons in the square planar, [ Pt ( CN )4]2- ion ?

Question

What is(are) number(s) of unpaired electrons in the square planar, [ Pt ( CN )4]2- ion ? (A) 0 (B) 1 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

In [ Pt ( CN )4]2-, CN -is a strong field ligand and outer configuration of Pt is 5 d9, 6 s1. Thus, outer configuration of Pt 2+ is 5 d8. CN -will cause pairing of electrons. Hence, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/405205e43a09b37ad65c431196e788b5-.png" /> Therefore, number of unpaired electrons in [ Pt ( CN )4]2- is zero.

Q. What is(are) number(s) of unpaired electrons in the square planar, $[Pt (CN)_{4}]^{2 -}$ ion ?

0

1

4

6

In $[Pt (CN)_{4}]^{2 -}, C N^{-}$ is a strong field ligand and outer configuration of Pt is $5 d^{9}, 6 s^{1}$ .
Thus, outer configuration of $P t^{2 +}$ is $5 d^{8}$ .
$C N^{-}$ will cause pairing of electrons. Hence,

Therefore, number of unpaired electrons in $[Pt (CN)_{4}]^{2 -}$ is zero.

Q. What is(are) number(s) of unpaired electrons in the square planar, [Pt(CN)4​]2− ion ?

0

1

4

6

In [Pt(CN)4​]2−,CN−is a strong field ligand and outer configuration of Pt is 5d9,6s1. Thus, outer configuration of Pt2+ is 5d8. CN−will cause pairing of electrons. Hence, Therefore, number of unpaired electrons in [Pt(CN)4​]2− is zero.

Q. What is(are) number(s) of unpaired electrons in the square planar, $[Pt (CN)_{4}]^{2 -}$ ion ?

In $[Pt (CN)_{4}]^{2 -}, C N^{-}$ is a strong field ligand and outer configuration of Pt is $5 d^{9}, 6 s^{1}$ .
Thus, outer configuration of $P t^{2 +}$ is $5 d^{8}$ .
$C N^{-}$ will cause pairing of electrons. Hence,

Therefore, number of unpaired electrons in $[Pt (CN)_{4}]^{2 -}$ is zero.