Which of the following statements is (are) incorrect?
(i) Percentage ionic character in `CsCl` bond is `~55%`. (Electronegativity values of Cs and Cl are 0.7 and 3.0 respectively)
(ii) From `H_(2)O` to `H_(3)O^(+)`, the geometry around O atom changes drastically .
(iii) The most stable oxidation state for element with atomic number 113 is expected to be `+2`.
(iv) The `2^("nd")` ionization energy of Ca is greater than `1^("st")` ionization energy of it but lower then `2^("nd")` ionization energyof K.

Options

A. Only i
B. i,ii, iii
C. ii, iii
D. iii, iv

Answer

To determine which of the statements are incorrect, let's analyze each statement step by step.

### Step 1: Analyze Statement (i)
**Statement:** Percentage ionic character in CsCl bond is ~55%. (Electronegativity values of Cs and Cl are 0.7 and 3.0 respectively)

**Solution:**
To calculate the percentage ionic character, we use the formula:
\[ \text{Percentage Ionic Character} = 16 \times (X_A - X_B) + 3.5 \times (X_A - X_B)^2 \]
Where \(X_A\) is the electronegativity of Cl (3.0) and \(X_B\) is the electronegativity of Cs (0.7).

Calculating the difference:
\[ X_A - X_B = 3.0 - 0.7 = 2.3 \]

Now substituting into the formula:
\[ \text{Percentage Ionic Character} = 16 \times 2.3 + 3.5 \times (2.3)^2 \]
Calculating:
\[ = 16 \times 2.3 + 3.5 \times 5.29 \]
\[ = 36.8 + 18.55 = 55.35\% \]

Thus, the statement is **correct**.

### Step 2: Analyze Statement (ii)
**Statement:** From H₂O to H₃O⁺, the geometry around the O atom changes drastically.

**Solution:**
- H₂O has a bent geometry due to two lone pairs of electrons on the oxygen atom, leading to a tetrahedral electron pair geometry but a bent molecular geometry.
- H₃O⁺ (hydronium ion) has three bond pairs and one lone pair, which also results in a tetrahedral electron pair geometry.

Both H₂O and H₃O⁺ have tetrahedral electron pair geometry. Therefore, the geometry does not change drastically.

Thus, this statement is **incorrect**.

### Step 3: Analyze Statement (iii)
**Statement:** The most stable oxidation state for element with atomic number 113 is expected to be +2.

**Solution:**
- Element 113 (Nihonium, Nh) has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p¹.
- Due to the inert pair effect, the outermost s-electrons (7s²) do not participate in bonding, leading to a more stable +1 oxidation state rather than +2.

Thus, this statement is **incorrect**.

### Step 4: Analyze Statement (iv)
**Statement:** The 2nd ionization energy of Ca is greater than the 1st ionization energy of it but lower than the 2nd ionization energy of K.

**Solution:**
- The first ionization energy of Ca (Ca → Ca⁺ + e⁻) is lower than that of K (K → K⁺ + e⁻) because K has a larger atomic radius and lower effective nuclear charge.
- The second ionization energy of Ca (Ca⁺ → Ca²⁺ + e⁻) is indeed higher than the first ionization energy of Ca.
- The second ionization energy of K (K⁺ → K²⁺ + e⁻) is higher than that of Ca because removing an electron from a noble gas configuration (after the first ionization) is more difficult.

Thus, this statement is **correct**.

### Conclusion
The incorrect statements are:
- Statement (ii) and Statement (iii).