Q 10P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Using the periodic table to locate each element, write the electron configuration of (a) V (b) Y (c) Hg.

Step-by-Step Solution

Answer

(a) The electron configuration of element V is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^3$or$\left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^3$.

 

(b) The electron configuration of element Y is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^1$or $\left[\mathrm{Kr}\right]5{\mathrm{s}}^{2}4{\mathrm{d}}^1$.

 

(c) The electron configuration of element Hg is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}$or $\left[\mathrm{Xe}\right]6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}3{\mathrm{d}}^{10}$.

Step 1: Electron Configuration

The distribution of electrons in atomic or molecular orbitals of an atom, molecule, or other physical structure is known as electron configuration.

Step 2: Electron configuration of element

(a) Based on the periodic table, V has the atomic number of $23$, therefore it has data-custom-editor="chemistry" $23$ electrons with electron configuration: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^3$ or $\left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^3$.

(b) Based on the periodic table, Y has the atomic number of $39$, therefore it has $39$ electrons with this electron configuration: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^1$ or $\left[\mathrm{Kr}\right]5{\mathrm{s}}^{2}4{\mathrm{d}}^1$.

 (c) Based on the periodic table, Hg has the atomic number of $80$, therefore it has $80$ electrons with this electron configuration: $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}$ or $\left[\mathrm{Xe}\right]6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}3{\mathrm{d}}^{10}$