Chapter 7, Problem 61QP

Interpretation Introduction

Interpretation:

The ions with three charges that have the given electronic configuration are to be named.

Concept introduction:

$\text{Ar}$ is a noble gas with the electronic configuration $\text{3}{s}^{2}3p^6$.

The $d$ block elements have the electronic configuration $(n-1)d^{1-10}n{s}^{0-2}$.

The last electron enters into any one of the five $d$ orbitals and not in the $s$ orbital.

Answer to Problem 61QP

Solution:

(a) ${\text{Cr}}^{3+}$

(b) ${\text{Sc}}^{3+}$

(c) ${\text{Rh}}^{3+}$

(d) ${\text{Ir}}^{3+}$

Explanation of Solution

a) $[\text{Ar}]3d^3$

Theoretically, the metal ion ${\text{M}}^{3+}$ having three electrons in the $3d$ subshell should have a total of sixvalence electrons and the element with that number should lie in group 6B. The transition metals tend to lose electrons from the ns valence subshell before the (n− 1) d subshell. There are only sixvalence electrons for a neutral atom. Group 6B elements have their ground state electronic configuration as $(n-1)d^{5}n{s}^1$, that is, ground state electronic configuration of the given atom should be $3d^{5}4s^1$. The element $\text{Cr}$ has this configuration and the ${\text{M}}^{3+}$ state as ${\text{Cr}}^{3+}$.

Hence, the metal ion ${\text{M}}^{3+}$ with the electronic configuration $[\text{Ar}]3d^3$ is ${\text{Cr}}^{3+}$.

b) $[\text{Ar}]$

$\text{Ar}$ is a noble gas with the electronic configuration $\text{3}{s}^{2}3p^6$. The metal ion ${\text{M}}^{3+}$ should have a total of threevalence electrons so as to attain $\text{Ar}$ configuration. The element with that number should lie in group 3B. The transition metals tend to lose electrons from the ns valence subshell before the (n− 1) d subshell. There are only threevalence electrons for a neutral atom. Group 3B elements have their ground state electronic configuration as $(n-1)d^{1}n{s}^2$, that is, ground state electronic configuration of the given atom should be $3d^{1}4s^2$. The element $\text{Sc}$ has this configuration and the ${\text{M}}^{3+}$ state having 18 electrons (as in ${\text{Ar}}_{18}$) is ${\text{Sc}}^{3+}$.

Hence, the metal ion ${\text{Sc}}^{3+}$ has the same electronic configuration as that of $[\text{Ar}]$.

c) $[\text{Kr}]4d^6$

$\text{Kr}$ is a noble gas with the electronic configuration $4s^{2}4p^6$. The metal ion ${\text{M}}^{3+}$ should have a total of 42electrons as $\text{Kr}$ has. Metal element with that number should lie in group 8B. Group 8B elements have their ground state electronic configuration as: $(n-1)d^{7/8/10}n{s}^{0-1}$. As per the question, the ground state electronic configuration of the given atom should be $4d^{8}5s^1$. The element ${\text{Rh}}_{45}$ has this configuration and the ${\text{M}}^{3+}$(i.e., 42 electrons) state as ${\text{Rh}}^{3+}$.

Hence, the metal ion ${\text{Rh}}^{3+}$ has the same electronic configuration as that of $[\text{Kr}]4d^6$.

d) $[\text{Xe}]4f^{14}5d^6$

$\text{Xe}$ is a noble gas with the electronic configuration $5s^{2}5p^6$. The metal ion ${\text{M}}^{3+}$ should have a total of 77 electrons. Metal element with that number should lie in group 8B. Group 8B elements have their ground state electronic configuration as: $(n-1)d^{7/8/10}n{s}^{0-1}$. As per the question, the ground state electronic configuration of the given atom should be $5d^{7}6s^1$. The element ${\text{Ir}}_{77}$ has this configuration and the ${\text{M}}^{3+}$(i.e.74 electrons) state as ${\text{Ir}}^{3+}$.

Hence, the metal ion ${\text{Ir}}^{3+}$ has the same electronic configuration as that of $[\text{Xe}]4f^{14}5d^6$.