Chapter 6, Problem 1KSP

Key Skills Problems

What is the noble gas core for Mo?

$\begin{array}{l}\left(\text{a}\right)\hfill \\ \text{Ar}\hfill \\ \left(\text{b}\right)\hfill \\ \text{Kr}\hfill \\ \left(\text{c}\right)\hfill \\ \text{Xe}\hfill \\ \left(\text{d}\right)\hfill \\ \text{Ne}\hfill \\ \left(\text{e}\right)\hfill \\ \text{Rn}\hfill \\ \hfill \end{array}$

Interpretation Introduction

Interpretation:

The noble gas core for $\text{Mo}$ is to be identified.

Concept introduction:

The number of protons present in the nucleusof an atom is equal to its atomic number.

The number of electrons in aneutral atom of an element is equal to its atomic number. With the help of thenumber of electrons, the electronic configuration of a given element can be deduced.

General rules for writing electronic configuration

Electrons will reside in the available orbitals of lowest possible energy.

Each orbital can accommodate a maximum of two electrons.

Electrons will not pair in degenerate orbitals if an empty orbital is available.

Orbitals will fill in order of increasing energy $1s2s2p3s3p4s3d4p5s4d5p6s4f$.

s-orbital can have 0–2 electrons.

p-orbital can have 0–6 electrons.

d-orbital can have 0–10 electrons.

f-orbital can have 0–14 electrons.

In the electronic configuration, the symbol of the noble gas in brackets represents the core electrons.

Answer to Problem 1KSP

Correct Answer: Option (b).

Explanation of Solution

Reasons for the correct option:

The atomic number of $\text{Mo}$ is $42$. The number of electrons present in a neutral atom of an element is equal to its atomic number. Thus, the number of electrons in $\text{Mo}$ isalso $42$. First, distribute two electrons in the $1s$ orbital, two electrons in the $2s$ orbital and six electrons in the $2p$ orbital. Further, distribute two electrons in the $3s$ orbital, six electrons in the $3p$ orbital, two electrons in the $4s$ orbital, ten electrons in the $3d$ orbital, six electrons in the $4p$ orbital, five electrons in the $4d$ orbital andone electron in the $5s$ orbital.

Hence, the electronic configuration of $\text{Mo}$ is as follows:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^5$

The inert gas that precedes molybdenum is krypton. So, the electron configuration for the core electrons is $\left[\text{Kr}\right]$, which have atomic number 36. Thus, the number of electrons in $\left[\text{Kr}\right]$ is 36.

Hence the electronic configuration of $\left[\text{Kr}\right]$ is as follows $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$

Hence, Option (b) is correct.

Reasons for the incorrect option:

Option (a) is incorrect because the total number of electrons in $\text{Mo}$ is 42. So, an inert gas that precedes molybdenum is krypton and the symbol for Kryptonis $\left[\text{Kr}\right]$. So, it is a wrong answer.

Option (c) is incorrect because the total number of electrons in $\text{Mo}$ is 42. So, an inert gas that precedes molybdenum is krypton and the symbol for Kryptonis $\left[\text{Kr}\right]$. So, it is a wrong answer.

Option (d) is incorrect because the total number of electrons in $\text{Mo}$ is 42. So, an inert gas that precedes molybdenum is krypton and the symbol for Kryptonis $\left[\text{Kr}\right]$. So, it is a wrong answer.

Option (e) is incorrect because the total number of electrons in $\text{Mo}$ is 42. So, an inert gas that precedes molybdenum is krypton and the symbol for Kryptonis $\left[\text{Kr}\right]$. So, it is a wrong answer.

Hence, options (a), (c), (d), and(e) are incorrect.