Chapter 2.9, Problem 52AAP

(a)

To determine

The electronic configuration of Cr²⁺, Cr³⁺ and Cr⁶⁺ using spdf notation.

Answer to Problem 52AAP

The electronic configuration of ${\text{Cr}}^{2+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$.

The electronic configuration of ${\text{Cr}}^{3+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$.

The electronic configuration of ${\text{Cr}}^{6+}$  ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

Explanation of Solution

There are certain number of sub-shells denoted by $\mathcal{l}$ that exist in any principal shell denoted by n. These sub shells are represented by the orbital quantum number. The quantum number is represented by an integer ranging from 0 to $n-1$ or by letters s, p, d, f and so on.

Thus, for $n=1$, $\mathcal{l}=s$; for $n=2$, $\mathcal{l}=s\text{or}p$ and so on.

The s sub-shell always looks spherical in shape but as the quantum number increases the size of the sphere increases which indicates that electrons can travel farther from the nucleus. However the p sub-shell is dumbbell shaped

The maximum number of electrons that can be contained in different sub-shells are $s=2$, $p=6$, $d=10$, $f=14$ and so on. The spdf configuration of electrons in any element is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$ and so on.

The element $\text{ Cr}$ has an atomic number 24 which means there are 24 electrons in $\text{ Cr}$ atom whereas ${\text{ Cr}}^{2+}$ ion contains 22 electrons.

Therefore the spdf configuration of ${\text{ Cr}}^{2+}$ ion is written as-$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$.

The number of electrons in ${\text{ Cr}}^{3+}$  is 21.

Therefore the electronic configuration of ${\text{Cr}}^{3+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$.

The number of electrons in ${\text{ Cr}}^{6+}$ ion is 18.

Therefore the electronic configuration of ${\text{ Cr}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

Thus the electronic configuration of ${\text{ Cr}}^{2+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$.

Thus, the electronic configuration of ${\text{ Cr}}^{3+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$.

Thus, the electronic configuration of ${\text{ Cr}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

(b)

To determine

The electronic configuration of ${\text{Mo}}^{\text{3+}}$, ${\text{Mo}}^{4+}$ and ${\text{Mo}}^{6+}$ using spdf notation.

Answer to Problem 52AAP

The electronic configuration of ${\text{Mo}}^{\text{3+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^3$.

The electronic configuration of ${\text{Mo}}^{4+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^2$.

The electronic configuration of ${\text{Mo}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^6$.

Explanation of Solution

The maximum number of electrons that can be contained in different sub-shells are $s=2$, $p=6$, $d=10$, $f=14$ and so on. The spdf configuration of electrons in any element is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$ and so on.

The atomic number of $\text{Mo}$ is 42 which means that an atom of $\text{Mo}$ contains 42 electrons whereas ${\text{Mo}}^{\text{3+}}$ contains three electrons less than that of $\text{Mo}$ which means that ${\text{Mo}}^{\text{3+}}$ contains 39 electrons.

Therefore, the electronic configuration of ${\text{Mo}}^{\text{3+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^3$.

The number of electrons in ${\text{Mo}}^{4+}$  is 38.

Therefore, the electronic configuration of ${\text{Mo}}^{4+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^2$.

The number of electrons in ${\text{Mo}}^{6+}$  is 36.

Therefore, the electronic configuration of ${\text{Mo}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^6$.

Thus, the electronic configuration of ${\text{Mo}}^{\text{3+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^3$.

Thus, the electronic configuration of ${\text{Mo}}^{4+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^2$.

Thus, the electronic configuration of ${\text{Mo}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^6$.

(b)

To determine

The electronic configuration of ${\text{Se}}^{\text{4+}}$, ${\text{Se}}^{6+}$ and ${\text{Se}}^{2-}$ using spdf notation.

Answer to Problem 52AAP

The electronic configuration of ${\text{Se}}^{\text{4+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$.

The electronic configuration of ${\text{Se}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$.

The electronic configuration of ${\text{Se}}^{2-}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$.

Explanation of Solution

The maximum number of electrons that can be contained in different sub-shells are $s=2$, $p=6$, $d=10$, $f=14$ and so on. The spdf configuration of electrons in any element is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$ and so on.

The atomic number of $\text{Se}$ is 21 which means that an atom of $\text{Se}$ contains 21 electrons whereas ${\text{Se}}^{\text{4+}}$ contains four electrons less than that of $\text{Se}$ which means that ${\text{Se}}^{\text{4+}}$ contains 17 electrons.

Therefore, the electronic configuration of ${\text{Se}}^{\text{4+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$.

The number of electrons in ${\text{Se}}^{6+}$  is 15.

Therefore, the electronic configuration of ${\text{Se}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$.

The number of electrons in ${\text{Se}}^{2-}$  is 23.

Therefore, the electronic configuration of ${\text{Se}}^{2-}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$.

Thus, the electronic configuration of ${\text{Se}}^{\text{4+}}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$.

Thus, the electronic configuration of ${\text{Se}}^{6+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$.

Thus, the electronic configuration of ${\text{Se}}^{2-}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$.