Class 11th

This is a Assertion Type Questions as classified in NCERT Exemplar

Ans: Option (i)

The statements given in the assertion and reason, both are correct and the reason given is the correct explanation for the assertion. In NH₃ only one lone pair of electron is present and due to that lone pair-bond pair repulsion occurs which forms a distorted tetrahedral geometry (pyramidal) .

In H₂O two lone pairs of electrons are present on the oxygen atom and due to this lone pairlone pair repulsion occurs which shifts the bonds and forms a bent geometry. The lone pair-lone pair repulsion is more than the lone pair-bond pair repulsion. Thus, bond angle in H-N-H is more than that of H-O-H.

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Tetrahedral    (c) sp³

(ii) Trigonal    (a) sp²

(iii) Linear    (b)sp

Explanation :-

(i) A tetrahedral molecule has four electron pairs and these make sigma bonds with each other. The s and p orbitals overlap with each other thus forming a sp³ hybridized molecule.

(ii) There are two possibilities for the central atom to have sp² Either all the bonds are in place i.e, the bond pairs are all sigma bonds or pi bonds or there are only two bonds and one lone pair of electrons.

(iii) The sp hybridization involves the mixing of the valence electrons in s orbital with another valence electron in the p orbital which yields two sp orbitals which are oriented in a linear geometry.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) Here, 'and' is the connecting word

The component statements are–

(a) All rational numbers are real

(b) All real numbers are not complex

(ii) Here, 'or' is the connecting word

The component statements are–

(a) Square of an integer is positive

(b) Square of an integer is negative

(iii) Here, 'and' is the connecting word

The component statements are–

(a) The sand heats up quickly in the Sun

(b) The sand does not caool down fast at night

(iv) Here, 'and' is the connecting word

The component statements are–

(a) n = 2 is the root of the equation 3x²–x– 10 = 0

(b) n = 3 is the root of the equation 3x²– x–10 = 0

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) Hydrogen bond                              (d) HF

(ii) Resonance    (e) O₃

(iii) Ionic bond                                    (b) LiF

(iv) Covalent solid    (a) C

Explanation :-

(i) In hydrogen fluoride HF, hydrogen forms strong hydrogen bonds with fluorine because of the electronegativity of fluorine. Fluorine is more electronegative than hydrogen and thus the electrons in fluorine are more stabilized and they cannot be easily donated to an acceptor.

(ii) Ozone is a neutral molecule which is stabilized through resonance. The negative charge created by an extra electron is delocalized by resonance through terminal oxygen atoms.

(iii) In lithium fluoride LiF, the lithium metal has a positive charge ( 1) + and is a cation. The fluoride ion has a negative charge ( 1) - and is an anion. These two oppositely charged ions are held together by an ionic bond.

(iv) Carbon is a known covalent solid. The two allotropes of carbon, diamond and graphite are covalent network solids with different geometries. Due to the presence of covalent bonds in graphite and diamond, they have very high melting points and are poor conductors of heat and electricity.

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) NO                                                    (c) 2.5

(ii) CO    (d) 3.0

(iii) O₂^2-                                                (a) 1.5

(iv) O₂                                                   (b) 2.0

Explanation :-

(i) The electronic configuration of NO is:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z² π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-5)= 2.5

(ii) The electronic configuration of CO is:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

The bond order will be:

$\frac{1}{2}$ ( 10-4)= 3.0

(iii) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-7)= 1.5

(iv) The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-6)= 2.0