Question

Which of the following reactions give phosphine? 
i. Reaction of calcium phosphide with water
ii. Heating white phosphorus with concentrated NaOH solution in an inert atmosphere
iii. Heating red phosphorus with alkali

• A. \( i, ii \text{ only} \)
• B. \( i, ii, iii \)
• C. \( ii, iii \text{ only} \)
• D. \( i, iii \text{ only} \)

Hint:

Phosphine is typically produced by hydrolysis of metal phosphides or by reduction of phosphorus compounds in basic conditions. However, red phosphorus does not react with alkali to give phosphine.

Answer 1

The Correct Option is A

The reactions that result in the formation of phosphine (PH₃) are analyzed below:

1. Reaction of calcium phosphide (Ca₃P₂) with water:
   This reaction yields phosphine and calcium hydroxide:
   \( \text{Ca}_3\text{P}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Ca(OH)}_2 + 2\text{PH}_3 \)
   Thus, option (i) produces phosphine.
2. Heating white phosphorus (P₄) with concentrated NaOH solution in an inert atmosphere:
   This reaction generates phosphine:
   \( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 3\text{NaH}_2\text{PO}_2 + \text{PH}_3 \)
   Therefore, option (ii) also leads to phosphine formation.
3. Heating red phosphorus with alkali:
   This reaction does not produce phosphine under normal conditions; instead, phosphite salts may form.
   Hence, option (iii) does not give phosphine.

From the options analyzed, the reactions that give phosphine are i and ii only.

The correct answer is: \( i, ii \text{ only} \)

Answer 2

Step 1: Understanding the nature of transition metal oxides Transition metals commonly form oxides in different oxidation states. The general formula \( MO \) is observed for several transition metals where the oxidation state of metal is \( +2 \). 
Step 2: Examining each option 
- Vanadium (V): Forms vanadium(II) oxide \( VO \).  
- Chromium (Cr): Forms chromium(II) oxide \( CrO \). 
- Manganese (Mn): Forms manganese(II) oxide \( MnO \). 
- Scandium (Sc): Does not form \( ScO \) because scandium primarily exists in the \( +3 \) oxidation state, leading to the formation of \( Sc_2O_3 \) rather than \( ScO \). 
Step 3: Conclusion Since scandium does not form \( MO \)-type oxides but instead forms \( Sc_2O_3 \), the correct answer is \( Sc \).