Choose the correct combination of true statements from the statements given below regarding `H_3BNH_3` and `H_(3)CCH_(3)`:
(i) the two molecules are isoelectronic
(ii) the two molecules are isostructural
(iii) both molecules have zero dipole moment
(iv) `H_3BNH_3` is paramagnetic while `H_3CCH_3` is diamagnetic
(v) both may be viewed as coordination compounds

Options

A. (i), (iii) and (iv)
B. (i) and (ii)
C. (ii) and (v)
D. (i), (ii) and (v)

Explanation

To solve the question regarding the compounds \( H_3BNH_3 \) (borazine) and \( H_3CCH_3 \) (ethane), we will analyze each statement one by one.

### Step 1: Determine if the two molecules are isoelectronic.
- **Isoelectronic Definition**: Two species are isoelectronic if they have the same number of total electrons.
- **Calculation**:
- For \( H_3BNH_3 \):
- Boron (B) has 5 electrons.
- Nitrogen (N) has 7 electrons.
- Each Hydrogen (H) has 1 electron, and there are 6 H atoms.
- Total electrons = \( 5 + 7 + (6 \times 1) = 18 \) electrons.

- For \( H_3CCH_3 \):
- Each Carbon (C) has 6 electrons, and there are 2 C atoms.
- Total electrons = \( (2 \times 6) + (6 \times 1) = 12 + 6 = 18 \) electrons.

- **Conclusion**: Both molecules have 18 electrons, so they are isoelectronic.

### Step 2: Determine if the two molecules are isostructural.
- **Isostructural Definition**: Two species are isostructural if they have the same structural arrangement.
- **Hybridization**:
- In \( H_3BNH_3 \), Boron and Nitrogen are both sp³ hybridized.
- In \( H_3CCH_3 \), both Carbon atoms are also sp³ hybridized.

- **Conclusion**: Since both compounds have the same hybridization (sp³), they are isostructural.

### Step 3: Check if both molecules have zero dipole moment.
- **Dipole Moment**:
- \( H_3BNH_3 \) has a dipole moment due to the difference in electronegativity between Boron and Nitrogen.
- \( H_3CCH_3 \) has no dipole moment because both Carbon atoms have the same electronegativity.

- **Conclusion**: Therefore, only \( H_3CCH_3 \) has zero dipole moment, while \( H_3BNH_3 \) does not. Hence, this statement is false.

### Step 4: Determine if \( H_3BNH_3 \) is paramagnetic while \( H_3CCH_3 \) is diamagnetic.
- **Magnetism**:
- \( H_3BNH_3 \) has no unpaired electrons and is diamagnetic.
- \( H_3CCH_3 \) also has no unpaired electrons and is diamagnetic.

- **Conclusion**: Both are diamagnetic, making this statement false.

### Step 5: Check if both may be viewed as coordination compounds.
- **Coordination Compound Definition**: A compound that contains coordinate covalent bonds.
- **Analysis**:
- \( H_3BNH_3 \) has a coordinate bond between Boron and Nitrogen.
- \( H_3CCH_3 \) does not have any coordinate bonds; it has covalent bonds.

- **Conclusion**: Only \( H_3BNH_3 \) can be classified as a coordination compound, making this statement false.

### Final Conclusion:
The correct combination of true statements regarding \( H_3BNH_3 \) and \( H_3CCH_3 \) is:
- (i) The two molecules are isoelectronic.
- (ii) The two molecules are isostructural.

Thus, the answer is statements (i) and (ii) are true.