Chemistry

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 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

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) H₃O⁺                                                                      (e) Pyramidal

(ii) HC? CH  (a) Linear

(iii) ClO₂^-                                                                    (b) Angular

(iv) NH4+                                                                    (c) Tetrahedral

Explanation :-

(i) The shape of H₃O⁺ is pyramidal this structure has the total 8 valence electrons, six in oxygen atom, three in each hydrogen atoms and one electron is lost due to positive charge. Out of five valence electrons in oxygen three form a sigma bond with the hydrogen atom and a lone pair is left on the oxygen atom. Although the four electron pairs make a tetrahedral geometry but to the lone pair of electrons, lone-pair bond-pair repulsion occurs which distorts the shape and makes it pyramidal.

(ii) The ethyne molecule has a linear geometry, because the two carbon atoms make two pi bonds with each other and one sigma bond with the hydrogen atom. The carbon and hydrogen atoms are sp hybridized.

(iii) ClO₂ ^- has 20 valence electrons in its structure. Chlorine atoms make one pibond and two sigma bonds with the oxygen atoms. The pi-bond is in resonance and hence ClO₂ ^- has two resonating structures. Two lone pairs of electrons are left on the oxygen at om and hence, due to lone pair-lone pair repulsion the bond angle reduces and the shape becomes linear.

(iv) NH₄⁺ has the total of 8 valence electrons. Due to the presence of positive charge, nitrogen has 4 valence electrons and hydrogen has one in every four hydrogen atoms. The four valence electrons in nitrogen make sigma bonds with the four hydrogen atoms, thereby making a tetrahedral geometry and having sp³ hybridized orbitals.

This is a Matching Type Questions as classified in NCERT Exemplar

Ans:

(i) SF₄    (c) sp³d

(ii) IF₅                                                       (a) sp³d²

(iii) NO₂⁺                                                 (e) sp

(iv) NH₄⁺                                                  (d) sp³

Explanation :-

(i) SF₄ has a total of 34 valence electrons. Sulphur has 6 valence electrons and each fluoride has 7 valence electrons. The central atom sulphur will thus share one electron with all the four fluoride ions and thus make four bonds. A lone pair of electrons (two electrons) is left on the sulphur atom. Thus, the sulphur atom is surrounded by five electron pairs, four bond pairs and one lone pair. Having five electron pairs surrounding sulphur, it will have a trigonal bipyramidal geometry with sp³d hybridization

(ii) IF₅ has the total of 42 valence shell electrons, 7 valence electrons in iodine and 7 in each of the fluoride ions. The central iodine atom will thus make five sigma bonds with chlorine and a lone pair of electrons will be left on the iodine atom. The six electron pairs thus will have a hybridization of sp³d² and octahedral geometry.

(iii) In NO₂^+, nitrogen forms two sigma bonds and two pi-bonds with the oxygen atom. The oxygen atoms are arranged in an angle of 180° with the nitrogen atoms. The s-orbital of the nitrogen forms a hybrid with the p-orbital of the nitrogen atom, thereby forming two sp hybridized hybrid orbitals, having a linear geometry.

(iv) NH₄⁺ has the total of 8 valence electrons. Due to the presence of positive charge, nitrogen has 4 valence electrons and hydrogen has one in every four hydrogen atoms. The four valence electrons in nitrogen make sigma bonds with the four hydrogen atoms, thereby making a tetrahedral geometry and having 3 sp hybridized orbitals.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and option (ii)

(i) NaCl being an ionic compound is a good conductor of electricity in the molten state. It is not a bad conductor of electricity in solid state. So, the statement mentioned in the question is not correct. Hence, this is the correct answer.

(ii) There is no difference in the arrangement of atoms in the canonical structures. The difference is in the arrangement of electron pairs in the canonical structures. So, the statement mentioned in the question is not correct. Hence, this is the correct answer.

(iii) The hybrid orbitals have the same energy and shape and hence they are more effective in the formation of stable bonds than the pure atomic orbitals. So, the statement given is correct. Hence, this is not the correct answer.

(iv) XeF₄ acquires a square planar geometry, this can very well be explained by the VSEPR theory. The hybridisation of the central xenon atom will be sp³d² due to the presence of two lone pairs of electrons on the central xenon atom. So, the statement given is correct. Hence, this is not the correct answer.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

The formula of bond order is:

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Number of bonding and antibonding electrons can be found from the molecular electronic configuration of the species.

(i) Total number of electrons in N2 is 14. So, its molecular electronic configuration will be

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

The bond order will be:

$\frac{1}{2}$ ( 10-4)= 3

(ii) The total number of electrons in N₂ ^- is 15. So, its molecular electronic configuration will be:

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

(iii) The total electrons in F₂⁺ is 17. So, its electronic configuration will be:

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 total electrons in O₂^- is 17 So, its electronic configuration will be

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

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (iv)

(i) N₂ will have 14 electrons in its structure. So, its electronic configuration will be:

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

According to the molecular electronic configuration of N₂ it does not contain any unpaired electrons, it will be diamagnetic in nature.

(ii) N₂^2– will have 16 electrons in its structure. So, its molecular electronic configuration will be:

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

According to the molecular electronic configuration of 2 N₂ ^-, it has two unpaired electrons in π*2p¹_x ,   π*2p¹_y  sub- shell. Hence, it will be paramagnetic in nature.

(iii) O₂ will have 16 electrons in its structure. So, its molecular electronic configuration will be:

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

According to the molecular electronic configuration of O₂, it has two unpaired electrons in π2p¹_y π*2p¹_x  sub-shell. Hence, it will be paramagnetic in nature.

(iv) O₂^2- will have 18 electrons in its structure. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p²_y π*2p²_x

According to the molecular electronic configuration of O₂^2- it does not contain any unpaired electrons, it will be diamagnetic in nature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

(i) The hybridization of the central carbon atom in CO₃ ^2- is 2 sp as the carbon is bonded with three oxygen atoms. So, the given statement is incorrect.

(ii) The resonance structure of CO₃^2- has one C= O double bonds and two C- O single bonds. So, the statement is incorrect.

(iii) The net charge on three oxygen atoms is -2, hence the average formal charge on each oxygen atom is 0.67 units. So, the given statement is correct.

(iv) As it can be seen from the resonating structures that the structures of the bonds are not fixed. Hence, all the C- O bond lengths are equal. So, the given statement is correct.