Find the correct statements related to group 15 hydrides. A. Reducing nature increases from NH3 to BiH3 B. Tendency to donate lone pair of electrons decreases from NH3 to BiH3 C. The stability of hydrides decreases from NH3 to BiH3 D. HEH bond angle decreases from NH3 to SbH3 (E = Elements of group 15) Choose the correct answer from the options given below :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q59 – JEE-MAIN 2026

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Home Chemistry Periodicity & ElementsQuestion #515

Find the correct statements related to group 15 hydrides. A. Reducing nature increases from

N H_{3}

B i H_{3}

B. Tendency to donate lone pair of electrons decreases from

N H_{3}

B i H_{3}

C. The stability of hydrides decreases from

N H_{3}

B i H_{3}

D. HEH bond angle decreases from

N H_{3}

S b H_{3}

(E = Elements of group 15) Choose the correct answer from the options given below :

Clue 1 of 3: Trends in Group 15 Hydrides

Recall the periodic trends for properties like reducing nature, basicity, stability, and bond angles as you move down a group.

🧠 Full Solution Path

Step 1: Analyze Reducing Nature and Stability✦ Active

Down Group 15, the atomic size of the central element (E) increases, leading to an increase in $E - H$ bond length and a decrease in bond strength. Weaker $E - H$ bonds mean easier release of hydrogen, thus increasing reducing nature (Statement A is correct). Weaker bonds also lead to lower thermal stability (Statement C is correct).

Step 2: Analyze Basicity (Lone Pair Donation)○ Expand

Down Group 15, the electronegativity of the central atom (E) decreases, and its size increases. This causes the lone pair of electrons to become more diffuse and less concentrated, making it less available for donation. Therefore, the tendency to donate a lone pair (Lewis basicity) decreases from $N H_{3}$ to $B i H_{3}$ (Statement B is correct).

Step 3: Analyze Bond Angle○ Expand

Down Group 15, the electronegativity of the central atom (E) decreases. This reduces the attraction of the bond pairs towards the central atom, causing them to occupy less space closer to the central atom. Consequently, bond pair-bond pair repulsion decreases, leading to a decrease in the $H - E - H$ bond angle. For example, bond angles are $N H_{3}$ ( $107^{\circ}$ ) > $P H_{3}$ ( ${93.5}^{\circ}$ ) > $A s H_{3}$ ( ${91.8}^{\circ}$ ) > $S b H_{3}$ ( ${91.3}^{\circ}$ ). Thus, the $H E H$ bond angle decreases from $N H_{3}$ to $S b H_{3}$ (Statement D is correct).

💡 Teacher's Secret Hint

Remember Drago's rule for heavier elements where hybridization might be less significant, leading to bond angles closer to $90^{\circ}$ .

Step 4: Conclusion○ Expand

Since all statements A, B, C, and D are correct, the correct option is 'A, B, C and D'.

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Chemistry Question 59 – JEE-MAIN 2026

Clue 1 of 3: Trends in Group 15 Hydrides

🧠 Full Solution Path