Download App

STD 11 - 4. Chemical bonding and molecular structure — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 4. Chemical bonding and molecular structure4 Marks
MCQ
Among the following ions the $p\pi  - d\pi $ overlap could be present in
  • A
    $NO_2^ - $
  • B
    $NO_3^ - $
  • ✓
    $PO_4^{ - 3}$
  • D
    $CO_3^{ - 2}$
✓

Answer

Correct option: C.
$PO_4^{ - 3}$
c
In $P=O$ bond, $\pi$ - bond is formed by the sidewise overlapping of $d$ - orbital of $P$ and $p$ - orbital of oxygen. Hence, it is formed by $p \pi$ and $d\pi$ overlapping.

In nitrogen and carbon, no vacant $d$-orbital is present. So, they do not form $p \pi- d \pi$ bond.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

NEET STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The melting point of $RbBr$ is $682\,^oC$ while that of $NaF$ is $988\,^oC$. The  principal reason for this fact is
→
To neutralise $20\, ml $ of $M/10$ sodium hydroxide, the volume of $M/20$ hydrochloric acid required is .............. $\mathrm{ml}$
→
Which of the following undergoes sulphonation at fastest rate ?
→
The aqueous solution of transition metal salt changes colour from pink to blue, when concentrated hydrochloric acid is added to it. The change In colour is due to
→
The number of elements present in fifth period is
→
The radii of $F,\,{F^ - },\,O$ and ${O^{ - 2}}$ are in the order of
→
Carbon mono-oxide is neutral oxide of carbon but it act as very strong field ligand and associated with metal to form complex compounds, these complex compounds are very useful compounds and uses for metallergical process. $CO$ is poisonous gas because :-
→
Given the correct order of initials $T$ or $F$ for following statements. Use $T$ if  statements is true and $F$ if it is false

$(I)\, (CH_3)_2P(CF_3)_3$ is non-polar and $(CH_3)_3P(CF_3)_2$ is polar  molecule
$(II)\, CH_3 \widehat{P} CH_3$ bond angles are equal in $(CH_3)_3P(CF_3)_2$ molecule
$(III)\,PF_3$ will be more soluble in polar solvent than $SiF_4$

→
The sugar present in milk is
→
Aniline on treatment with conc. $HN{O_3} + $ conc. ${H_2}S{O_4}$ mixture yields
→