Q.Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.

Concept and Intuition

The key idea here is orbital energy vs. nuclear charge. For a hydrogen-like atom (one electron), the energy of an orbital depends only on the principal quantum number $n$ and the nuclear charge $Z$:

But for multi-electron atoms, things get more complex due to electron-electron repulsion and shielding. However, the trend is clear: as atomic number increases, the effective nuclear charge felt by an electron in a given orbital increases, pulling it closer and making its energy more negative (lower).

• Hydrogen ($Z=1$): The 2s electron feels the full +1 charge.
• Lithium ($Z=3$): The 2s electron is shielded by the two 1s electrons, but the effective nuclear charge is still >1 (roughly +1.3). So the 2s electron in Li is more tightly bound than in H.

Thus, the energy of the 2s orbital in Li is lower (more negative) than in H. The assertion says the opposite — so it is incorrect.

The reason states that energies of orbitals in the same subshell decrease with increasing atomic number — this is generally true for multi-electron atoms (and even for hydrogen-like atoms if we compare same $n$ but different $Z$). So the reason is correct.

Step-by-step reasoning

1. Understand the assertion

   Assertion: "Energy of 2s orbital of Hydrogen is less than the energy of the 2s orbital of Lithium."

   "Less" here means more negative (lower energy). So it claims $E_{2s}(\text{H}) < E_{2s}(\text{Li})$.

2. Recall orbital energy dependence

   For a hydrogen atom (single electron), $E_{2s} = -13.6 / 2^2 = -3.4\ \text{eV}$.

   For lithium, the 2s electron experiences an effective nuclear charge $Z_{\text{eff}} \approx 1.3$ (due to shielding by 1s²). Using the hydrogen-like formula as an approximation:

So $-5.75\ \text{eV} < -3.4\ \text{eV}$, meaning the 2s orbital of Li is lower in energy than that of H.

Therefore, the assertion is false.

3. Evaluate the reason Reason: "Energies of orbitals in the same subshell decrease with increase in atomic number." …

The question tests your understanding of how bond order and bond length relate in diatomic molecules, specifically for oxygen and its ions. Bond order is directly linked to stability and inversely linked to bond length: higher bond order means a shorter, stronger bond. For oxygen species, we use the molecular orbital (MO) diagram for $\mathrm{O}_2$, which has 16 electrons in the neutral molecule. The MO configuration for $\mathrm{O}_2$ is:

The bond order formula is: $\text{Bond order} = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons})$. For $\mathrm{O}_2$, bonding electrons = 10 (from $\sigma_{2s}$, $\sigma_{2p_z}$, $\pi_{2p_x}$, $\pi_{2p_y}$ — careful: $\sigma_{2s}$ is bonding, $\sigma_{2s}^*$ is antibonding), and antibonding electrons = 6, giving bond order = 2. Now let's evaluate each species.

1. Determine electron counts and MO configurations for each species.

   • $\mathrm{O}_2$: 16 electrons. Bond order = $\frac{10 - 6}{2} = 2$.
   • $\mathrm{O}_2^+$: remove 1 electron from $\mathrm{O}_2$. The electron removed is from a $\pi^*$ orbital (highest occupied). So bonding = 10, antibonding = 5. Bond order = $\frac{10 - 5}{2} = 2.5$.
   • $\mathrm{O}_2^-$: add 1 electron to $\mathrm{O}_2$, which goes into a $\pi^*$ orbital. Bonding = 10, antibonding = 7. Bond order = $\frac{10 - 7}{2} = 1.5$.
   • $\mathrm{O}_2^{2-}$: add 2 electrons to $\mathrm{O}_2$, both into $\pi^*$ orbitals. Bonding = 10, antibonding = 8. Bond order = $\frac{10 - 8}{2} = 1.0$.
   • $\mathrm{O}_2^{2+}$: remove 2 electrons from $\mathrm{O}_2$. Both come from $\pi^*$ orbitals. Bonding = 10, antibonding = 4. Bond order = $\frac{10 - 4}{2} = 3.0$.

2. Analyze each statement.

   • (A) Bond length of $\mathrm{O}_2 >$ Bond length of $\mathrm{O}_2^{2+}$. Bond order of $\mathrm{O}_2$ = 2, bond order of $\mathrm{O}_2^{2+}$ = 3. Higher bond order means shorter bond length, so $\mathrm{O}_2$ has a longer bond than $\mathrm{O}_2^{2+}$. This is correct. …

Step 1 — Distinguish the two kinds of hardness

• Temporary hardness — due to dissolved bicarbonates of Ca and Mg: $Ca(HCO_3)_2$, $Mg(HCO_3)_2$. Removed simply by boiling, or by Clark's method.
• Permanent hardness — due to the chlorides and sulphates of Ca and Mg: $CaCl_2$, $MgCl_2$, $CaSO_4$, $MgSO_4$. Boiling does nothing; the $Ca^{2+}/Mg^{2+}$ must be chemically removed or exchanged.

Step 2 — Check each method against permanent hardness

(A) Washing soda ($Na_2CO_3\cdot10H_2O$) — the carbonate ion precipitates the offending cations as insoluble carbonates:

Works. ✓

(B) Calgon's method — sodium hexametaphosphate, $Na_6P_6O_{18}$ (written $Na_2[Na_4(PO_3)_6]$), sequesters $Ca^{2+}/Mg^{2+}$ into a soluble complex anion, so they can no longer form scum:

Works. ✓

(C) Clark's method — a calculated quantity of slaked lime, $Ca(OH)_2$, is added. Its action is on bicarbonates:

$Ca(HCO_3)_2 + Ca(OH)_2 \longrightarrow 2CaCO_3\!\downarrow + 2H_2O$ …

Step 1 — What $\mathrm{IE_3}$ actually measures.

So the decisive question is: how stable is the $M^{2+}$ ion? The more stable (closed-shell) it is, the larger $\mathrm{IE_3}$.

Step 2 — Examine each family.

• Alkaline earth metals (group 2, $ns^2$): $M^{2+}$ has the configuration of the preceding noble gas, e.g. $\mathrm{Mg^{2+}} = [\mathrm{Ne}]$, $\mathrm{Ca^{2+}} = [\mathrm{Ar}]$. Removing a third electron breaks into a complete octet and does so against a $+2$ nuclear pull ⇒ very large $\mathrm{IE_3}$.
• Alkali metals (group 1, $ns^1$): their second IE is the huge one (that is the one that breaks the noble-gas core). By $\mathrm{IE_3}$ the ion is already $M^{2+}$ = (noble gas $-\,1$ electron), i.e. an incomplete shell — easier to ionise than a complete one at the same period.
• Chalcogens ($ns^2np^4$) and pnictogens ($ns^2np^3$): their $M^{2+}$ ions still have $p$-electrons in the valence shell, which are comparatively easy to remove ⇒ modest $\mathrm{IE_3}$.

Step 3 — Confirm with data (kJ mol$^{-1}$), comparing within a period. …

Step 1 — Place the elements.

• Period 2: C (group 14), N (group 15)
• Period 3: Si (group 14), P (group 15)

Step 2 — Trend down a group. Ionisation enthalpy decreases down a group: the outermost electron is farther from the nucleus and better shielded, so it is easier to remove.

Step 3 — Trend across a period. Ionisation enthalpy increases across a period (nuclear charge increases, size decreases):

Step 4 — Combine, using the actual values. These trends alone leave the C-vs-P comparison open (C is up-and-left of P), so use the data:

Step 1 — Compare H$^-$, H and H$^+$ (same nucleus, different electron counts).

• $\mathrm{H^{+}}$ is a bare proton — no electron cloud at all. Its radius is of nuclear order ($\sim 10^{-15}$ m), vanishingly small compared with atomic radii ($\sim 10^{-10}$ m).
• $\mathrm{H}$ has 1 proton, 1 electron.
• $\mathrm{H^{-}}$ has 1 proton but 2 electrons. The same single proton must now hold two electrons, so the nuclear charge per electron drops, inter-electronic repulsion rises, and the cloud expands.

Therefore $\ \mathrm{H^{-} > H > H^{+}}$ — option (A) is consistent.

Step 2 — Check the other options (they are all isoelectronic sets).

For isoelectronic species (same number of electrons), the rule is: radius decreases as nuclear charge $Z$ increases — more protons pull the same number of electrons in more tightly. Equivalently, size increases as the negative charge increases. …