In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?
- CO2 < SiO2 < SnO2 < PbO2 : Increasing oxidizing order
- HF < HCl < HBr < HI : Increasing acidic strength
- H2O < H2S < H2Se < H2Te : Increasing pKa values
- NH3 < PH3 < AsH3 < SbH3 : Increasing acidic character
Answer
(3) H2O < H2S < H2Se < H2Te : Increasing pKa values
Solution
(1) Down group 14 the +2 state becomes more stable (inert pair effect), so PbO2 is the strongest oxidising agent ⇒ order correct.
(2) Down group 17 the H–X bond enthalpy decreases ⇒ acidic strength HF < HCl < HBr < HI ⇒ correct.
(3) As per NCERT, the acidic character of hydrides of group 16 increases from H2O to H2Te (H–E bond enthalpy decreases). Stronger acid ⇒ larger Ka ⇒ smaller pKa.
∴ pKa decreases: H2O > H2S > H2Se > H2Te ⇒ the given sequence is wrong.
(4) Down group 15 the E–H bond becomes weaker ⇒ acidic character increases NH3 < PH3 < AsH3 < SbH3 ⇒ correct.
Answer: (3)