NEET

The strongest hydrogen bond is :

O – H...........S
S – H.............O
F – H............F
F – H.............O

Solution

Higher the electronegativity of the other atom, greater is the strength of hydrogen bond. Strongest hydrogen bond is between H and F.
F – H ------- F.

The number of possible resonance structures for \(CO_{3}^{2-}\) is

2
3
6
9

Solution

There are three resonance structures of \(CO_{3}^{2-}\) ion.

How many sigma bonds are there in P₄O₁₀?

4
8
12
16

Solution

The structure of P₄O₁₀ is

The number of σ bonds in it =16
[Note:Single bonds are σ-bonds. Double bond consists of 1σ and 1π bond].

Among the following the electron deficient compound is

BCl₃
CCl₄
PCl₅
BeCl₂

Solution

Boron in BCl₃ has 6 electrons in outermost shell. Hence BCl₃ is a electron deficient compound.

As the s-character of hybridised orbital increases, the bond angle

increase
decrease
becomes zero
does not change

Solution

Bond angle increases with increase in s-character of hybridised orbital. The table given below shows the hybridised orbitals, their % s-chatracter and bond angles.

Which of the following hydrogen bonds are strongest invapour phase?

HF--HF
HF--HCl
HCl--HCl
HF--HI

Solution

A compound having element with highest electronegativity will form strongest hydrogen bond.

In PO\(PO_{4}^{3-}\) ion, the formal charge on each oxygen atom and P—O bond order respectively are

–0.75, 0.6
– 0.75, 1.0
– 0.75, 1.25
–3, 1.25

Solution

Bond order between P– O

\(=\frac{no.\: of\: bonds\: in\: all\: possible\: direction}{total\: no.\: of\: resonating\: structures}\)

=⁵⁄₄=1.25

Formal charge on oxygen=-³⁄₄=-0.75

What is the correct mode of hybridisation of the centralatom in the following compounds?
NO₂– SF₄ PF₆–

sp sp² sp³
sp² sp³d sp³d²
sp² sp³d d²sp³
sp³ sp³d d²sp³

Solution

Hybridisation of the central atom in compound is given by
H =¹⁄₂[V+M-C+A]
where V = no.of valency electrons in central metalatom,
M = no. of monovalent atoms surrounding the centralatom,
C = charge on cation and A = charge on anion
•For NO₂-,H = ¹⁄₂[5+0-0+1]=3
sp² hybridisation
•For SF₄, H =¹⁄₂[6+4-0+0]=5
sp³d hybridisation
•For PF₆–, H = ¹⁄₂[5+6-0+1]=6
sp³d²hybridisation.
So, option (a) is correct choice.

Indicate the nature of bonding in CCl₄ and CaH₂

Covalent in CCl₄ and electrovalent in CaH₂
Electrovalent in both CCl₄ and CaH₂
Covalent in both CCl₄ and CaH₂
Electrovalent in CCl₄ and covalent in CaH₂

When a metal atom combines with non-metal atom,the non-metal atom will

lose electrons and decrease in size
lose electrons and increase in size
gain electrons and decrease in size
gain electrons and increase in size

Solution

When a metal for example Nacombines with anonmetal e.g., Cl₂. Following reaction occurs
2NaCl₂→NaCl
In this process Na loses one electron to form Na⁺ and Cl accepts one electron to form Cl
Na→Na⁺+e^-
Cl+e^-→Cl^-
Therefore,in this process Cl gain electrons and hence its size increases.

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