Geometrical shapes of the complexes (p),(q),(r) formed by the following reactions respectively are :
`Cu^(2+)(aq)+KCN(aq)("excess")to"complex"(p)`
`Co^(2+)(aq)+KNO_2(s)+H^+ (aq)to"complex"(q)`
`Zn^(2+)(aq)+NaOH(aq)("excess")to"complex"(r)`

Options

A. Tetrahedral , octahedral and square planar
B. tetrahedral , octahedral and tetrahedral
C. Square planar , octahedral and tetrahedral
D. Octahedral, octahedral and tetrahedral

Explanation

To determine the geometrical shapes of the complexes formed in the given reactions, we will analyze each reaction step-by-step.

### Step 1: Analyze the first reaction
**Reaction:**
\[ \text{Cu}^{2+}(aq) + \text{KCN}(aq) \text{ (excess)} \rightarrow \text{complex (p)} \]

**Formation of the complex:**
- Copper(II) ion (\(\text{Cu}^{2+}\)) reacts with cyanide ions (\(\text{CN}^-\)).
- The complex formed is \(\text{Cu(CN)}_4^{3-}\).

**Oxidation state of Copper:**
- Let the oxidation state of Cu be \(x\).
- The total charge from 4 cyanide ions is \(-4\).
- The overall charge of the complex is \(-3\).
- Therefore, \(x - 4 = -3\) leads to \(x = +1\).

**Electron configuration of Cu:**
- The electron configuration of \(\text{Cu}\) is \(4s^1 3d^{10}\).
- For \(\text{Cu}^+\), it becomes \(4s^0 3d^{10}\) (fully filled d-orbitals).

**Hybridization:**
- The complex has 4 ligands (CN\(^-\)), which leads to \(sp^3\) hybridization.

**Geometry:**
- \(sp^3\) hybridization corresponds to a tetrahedral geometry.

### Conclusion for complex (p):
**Shape:** Tetrahedral

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### Step 2: Analyze the second reaction
**Reaction:**
\[ \text{Co}^{2+}(aq) + \text{KNO}_2(s) + \text{H}^+(aq) \rightarrow \text{complex (q)} \]

**Formation of the complex:**
- Cobalt(II) ion (\(\text{Co}^{2+}\)) reacts with nitrite ions (\(\text{NO}_2^-\)) and protons (\(H^+\)).
- The complex formed is \(\text{Co(NO}_2)_6^{3-}\).

**Oxidation state of Cobalt:**
- Let the oxidation state of Co be \(x\).
- The total charge from 6 nitrite ions is \(-6\).
- Therefore, \(x - 6 = -3\) leads to \(x = +3\).

**Electron configuration of Co:**
- The electron configuration of \(\text{Co}\) is \(4s^2 3d^7\).
- For \(\text{Co}^{3+}\), it becomes \(4s^0 3d^6\).

**Hybridization:**
- The complex has 6 ligands (NO\(_2^-\)), leading to \(d^2sp^3\) hybridization.

**Geometry:**
- \(d^2sp^3\) hybridization corresponds to an octahedral geometry.

### Conclusion for complex (q):
**Shape:** Octahedral

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### Step 3: Analyze the third reaction
**Reaction:**
\[ \text{Zn}^{2+}(aq) + \text{NaOH}(aq) \text{ (excess)} \rightarrow \text{complex (r)} \]

**Formation of the complex:**
- Zinc(II) ion (\(\text{Zn}^{2+}\)) reacts with hydroxide ions (\(\text{OH}^-\)).
- The complex formed is \(\text{Zn(OH)}_4^{2-}\).

**Oxidation state of Zinc:**
- Let the oxidation state of Zn be \(x\).
- The total charge from 4 hydroxide ions is \(-4\).
- Therefore, \(x - 4 = -2\) leads to \(x = +2\).

**Electron configuration of Zn:**
- The electron configuration of \(\text{Zn}\) is \(4s^2 3d^{10}\).
- For \(\text{Zn}^{2+}\), it becomes \(4s^0 3d^{10}\) (fully filled d-orbitals).

**Hybridization:**
- The complex has 4 ligands (OH\(^-\)), leading to \(sp^3\) hybridization.

**Geometry:**
- \(sp^3\) hybridization corresponds to a tetrahedral geometry.

### Conclusion for complex (r):
**Shape:** Tetrahedral

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### Final Summary of Complex Shapes:
- **Complex (p):** Tetrahedral
- **Complex (q):** Octahedral
- **Complex (r):** Tetrahedral