Question

An ion having a $\mathbf{4}\mathbf{+}$charge and a mass of $\mathbf{49}\mathbf{.}\mathbf{9}$amu, has two electrons with $\mathit{n}\mathbf{=}\mathbf{1}$, eight electrons with $\mathit{n}\mathbf{=}\mathbf{2}$and ten electrons with $\mathit{n}\mathbf{=}\mathbf{3}$. Supply the following properties for the ion (Hint. In the following ions the us electrons are lost before $\mathbf{3}\mathit{d}\mathbf{-}$electrons)

(a) the atomic number

(b) total number of $\mathit{s}\mathbf{-}$electrons

(c) total number of $\mathit{p}\mathbf{-}$electrons

(d) total number of $\mathit{d}\mathbf{-}$electrons

(e) The number of neutrons in the nucleus

(f) the mass of $\mathbf{3}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$atoms

(g) the ground-state electron configuration of the neutral atom.

Short answer

Answer:

Ion: An atom or molecule that has gained or lost one or more electrons and so has a positive or negative electric charge.

Ex:

$H\to H^{+}+e$

$\frac{1}{2}Cl+e^{-}=C{l}^{-}$

Step-by-step solution

Principal Quantum Number:

It is designated by symbol $n$and it describes the energy of an electron and the most probable distance of the electron from the nucleus, it states about the principle shell associated with that electron.

Angular quantum number values vary from n to (N-1):

$$\begin{gathered} For\mathit{}n\mathit{=}\mathit{1}\mathit{,}l\mathit{=}\mathit{0}\mathit{,}s\mathit{-}subshell \\ n\mathit{=}\mathit{2}\mathit{,}l\mathit{=}l\mathit{,}p\mathit{-}subshell \\ n\mathit{=}\mathit{3}\mathit{,}l\mathit{=}\mathit{2}\mathit{,}d\mathit{-}subshell \\ n\mathit{=}\mathit{4}\mathit{,}l\mathit{=}\mathit{3}\mathit{,}f\mathit{-}subshell \end{gathered}$$

Subpart (a): the atomic number: 

Step 1:

Given

2 electron in$n=1$

8 electron with$n\mathit{=}\mathit{2}$

10 electron with$n\mathit{=}\mathit{3}$

And having $\mathit{+}\mathit{4}$charge and Mass $\mathit{=}\mathit{49}\mathit{.}\mathit{9}amu$

Step 2:

$M^{\mathit{4}\mathit{+}}$ion total number of electrons $\mathit{=}\mathit{2}$

Electronic configuration$\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^{\mathit{2}}$

Number of electrons in $$\begin{gathered} M\mathit{=}\mathit{20}\mathit{+}\mathit{4} \\ \mathit{}\mathit{}\mathit{}\mathit{=}\mathit{24} \end{gathered}$$

$\therefore$No. of electron in an atom = No of proton = Atomic No = 24

Hence an atomic number of the ion is =24 .

Subpart (b): Total number of s- electrons:

Step 1:

Electronic configuration of the ion $\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^{\mathit{2}}$

Step 2:

Total number of $s^{\mathit{-}}$electron$\mathit{=}\mathit{2}\mathit{+}\mathit{2}\mathit{+}\mathit{2}\mathit{+}\mathit{=}\mathit{6}$

Subpart (c): Total number of  electrons:

Step 1:

Electronic configuration of the ion $\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^{\mathit{2}}$

Step 2:

Total number of $p^{\mathit{-}}$electron$\mathit{6}\mathit{+}\mathit{6}\mathit{=}\mathit{12}$

Subpart (d): Total number of  d- electrons:

Step 1:

Electronic configuration of the ion$\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^{\mathit{2}}$

Step 2:

Total number of $d^{\mathit{-}}$electron$\mathit{=}\mathit{2}$

Subpart (e): The number of neutrons in the nucleus:

Step 1:

The number of neutrons $\mathit{=}A\mathit{-}Z$

= Mass Number - Atomic Number

Step 2:

The atomic number of ion$\mathit{=}\mathit{24}$

The mass number of ion$\mathit{=}\mathit{49}\mathit{.}\mathit{9}$

Therefore the number of neutrons$\mathit{=}\mathit{49}\mathit{.}\mathit{9}\mathit{-}\mathit{24}\mathit{=}\mathit{25}\mathit{.}\mathit{9}$

Subpart (f): The mass of  3.01×1023atoms:

Mass of ion $=49.9\mathrm{amu}=49.9\times 1.66\times {10}^{-24}\mathrm{g}$

$\mathit{=}\mathit{82}\mathit{.}\mathit{83}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{24}}g$

And Avogadro’s number$\mathit{=}\mathit{6}\mathit{.}\mathit{022}\mathit{\times }{\mathit{10}}^{\mathit{23}}mo{l}^{\mathit{-}\mathit{1}}$

We know, Number of moles $\mathit{=}\frac{\mathit{N}}{{\mathit{N}}_{\mathit{A}}}$

where $N\mathit{=}a$a number of particles and $N_{\mathit{a}}\mathit{=}$Avogadro’s Number

$$\begin{gathered} n\mathit{=}\frac{\mathit{3}\mathit{.}\mathit{01}\mathit{\times }{\mathit{10}}^{\mathit{23}}}{\mathit{6}\mathit{.}\mathit{02}\mathit{\times }{\mathit{10}}^{\mathit{23}}} \\ \mathit{}\mathit{}\mathit{=}\frac{\mathit{1}}{\mathit{2}}mole \end{gathered}$$

$\mathit{\therefore }n\mathit{=}\frac{\mathit{W}}{\mathit{M}}$[Weight or mass in $g$and $M\mathit{=}$Molecular or atomic mass in $g$]

$\mathit{\therefore }\frac{\mathit{1}}{\mathit{2}}\mathit{=}\frac{W}{\mathit{82}\mathit{.}\mathit{83}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{24}}g}$

Therefore, $W\mathit{=}\frac{\mathit{82}\mathit{.}\mathit{83}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{24}}g}{\mathit{2}}$

$\mathit{}\mathit{=}\mathit{41}\mathit{.}\mathit{415}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{24}}g$

Hence $\mathit{=}\mathit{41}\mathit{.}\mathit{415}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{24}}g$mass

Subpart (g): The ground-state electron configuration of the neutral atom:

$M^{\mathit{4}\mathit{+}}$has electronic configuration $\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^{\mathit{2}}$

$\mathit{\therefore }M$has electronic configuration $\mathit{=}\mathit{1}{s}^{\mathit{2}}\mathit{,}\mathit{2}{s}^{\mathit{2}}\mathit{2}{p}^{\mathit{6}}\mathit{,}\mathit{3}{s}^{\mathit{2}}\mathit{3}{p}^{\mathit{6}}\mathit{3}{d}^4\mathit{,}\mathit{4}{s}^{\mathit{2}}$