Q 12P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Using the periodic table to locate each element, write theelectron configuration of (a) Os (b) Co (c) Ag.

Step-by-Step Solution

Answer

1. Electron Configuration of element Os is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^7$ or $\left[\mathrm{Xe}\right]6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^6$.
2. Electron Configuration of element Co is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^7$ or $\left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^7$.
3. Electron Configuration of element Ag is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{1}4{\mathrm{d}}^{10}$or $\left[\mathrm{Kr}\right]5{\mathrm{s}}^{1}4{\mathrm{d}}^{10}$.

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, Os has the atomic number of$76$, therefore it has $76$ electrons with electron configuration:

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^7$ or $\left[\mathrm{Xe}\right]6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^6$.

 

(b) Based on the periodic table, Co has the atomic number of $27$, therefore it has $27$ electrons with this electron configuration:

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^7$ or $\left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^7$.

 

(c) Based on the periodic table, Ag has the atomic number of $47$, therefore it has $47$ electrons with electron configuration:

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{1}4{\mathrm{d}}^{10}$or $\left[\mathrm{Kr}\right]5{\mathrm{s}}^{1}4{\mathrm{d}}^{10}$.

Instead of having a full 5s orbital, the atom is more stable when the d-orbital is fully filled.