Chemistry NCERT Exemplar Solutions Class 11th Chapter Three: Overview, Questions, Preparation

This is a Long Type Questions as classified in NCERT Exemplar

The factors that affects electron gain enthalpy are as follows:

(a) Effective nuclear charge with its increase the electron gain enthalpy also increases

(b) Atomic size with its increase the electron gain enthalpy decreases

(c) The e-e - repulsion with its increase the electron gain enthalpy decreases

(d) Whenever there exists half filled or completely filled orbital than it lead to decrease the electron gain enthalpy because it leads to give extra stability to the atom due to symmetry

The electron gain enthalpy decreases down the group and increases across the period.

This is a Long Type Questions as classified in NCERT Exemplar

The amount of energy required to remove an electron from the outermost shell of an isolated gaseous atom in its ground state is said to be ionisation enthalpy.

The factors that affect ionization enthalpy of the elements are as follows:

(a) Effective nuclear charge with its increase the ionization enthalpy also increases

(b) Atomic size with its increase the ionization enthalpy decreases

(c) The e-e - repulsion with its increase the ionization enthalpy decreases

(d) Whenever there exists half filled or completely filled orbital than it lead to increase the ionization enthalpy because it leads to give extra stability to the atom due to symmetry

The ionization enthalpy decreases down the group and increases across the period.

This is a Long Type Questions as classified in NCERT Exemplar

The chemical properties and physical properties of the elements depends upon its outer electron configuration which ultimately is the periodic function of the atomic number. The elements which possess the same outer electron configuration belong to the same group or family showing similar properties.

This is a Long Answer Type Questions as classified in NCERT Exemplar

The group 1 elements are called alkali metals which possess outermost electronic configuration as ns1 .All the elements that have same outermost electron configuration possess similar properties and are placed in the same group.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Mendeleev's arranged the elements as the periodicity of their atomic weights.

The drawbacks of Mendeleev's periodic table are as follows:-

(a) The position of hydrogen in the periodic table is not specified

(b) Isotopes are not included in the periodic table

(c) Elements with higher atomic mass are placed before the elements with lower atomic mass. For e.g- Co & Ni

(d) Gaps are left in his table considering the fact that more elements are yet to be discovered

(e) Inappropriate position of group VII.

This is a Long Answer Type Questions as classified in NCERT Exemplar

The following points make long form of periodic table better than Mendeleev's periodic table :-

(a) It is a periodic function of atomic number

(b) Elements are grouped as per there outermost electronic configuration

(c) Proper segregation of metals and non-metals

(d) More appropriate position of group VII.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: As we move across the periodic table the ionization energy increases because of the increase in the effective nuclear charge and decrease of the shielding effect as more and more electrons get added in the same orbital.

Thus group 1 has lower ionization enthalpy compared to that of group 17 and also group 1 by losing one electron it will acquire the nearest noble gas electronic configuration which also contributes towards its lower ionization enthalpy.

This is a Short Answer Type Questions as classified in NCERT Exemplar

As the atomic size of F is much smaller which leads to high e^- - e ^- repulsion upon addition of electrons thus its electron gain enthalpy is less than that of Cl.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Transition elements are named so because they form a bridge between s-block elements and p-block elements. Zn, Cd and Hg are among those elements that are dblock elements but they do not exhibit most of the properties of transition elements.

This is a Short Answer Type Questions as classified in NCERT Exemplar

This is a Short Answer Type Questions as classified in NCERT Exemplar

This is a Short Answer Type Questions as classified in NCERT Exemplar

(i) C has the highest ionization energy among the given elements as along the period ionization enthalpy increases whereas it decreases down the group.

(ii) Al is the most metallic element among the given elements because down the group metallic character increases

This is a Short Answer Type Questions as classified in NCERT Exemplar

The characteristics properties of p-block elements are as follows:-

(i) It contain metals, nonmetals and metalloids

(ii) Mostly involved in covalent bonding

(iii) Some elements show variable oxidation state

(iv) It possesses relatively higher ionization enthalpy compared to the s-block elements.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Option (i)

As across the period the atomic radius decreases due to the increase of effective nuclear charge.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The electronic configuration of Cr= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰

The electronic configuration of Cr after losing one electron= 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

The electronic configuration of F= 1s² 2s² 2p⁶ 3s² 3p⁵

The electronic configuration of F after gaining one electron= 1s² 2s² 2p⁶ 3s² 3p⁶

From the above electronic configuration, we can see that chromium will achieve stable electronic configuration after losing one 4s electron and fluorine will achieve stable electronic configuration after gaining one electron. So, the oxidation of chromium will be +1 and that of fluorine will be -1.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The N possesses a half filled p-orbital which provides it extra stability due to symmetry due to which its electron gain enthalpy is positive and its ionization enthalpy is larger than that of O.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The anomalous behaviour of the first member of each group of representative elements i.e. of second period can be attributed to their small size, high charge/radius ratio, high electronegativity and absence of vacant d-orbitals to expand their oxidation state.The first member of each group of p-Block elements displays their greater ability to form multiple bond with itself, e.g. C=C, O=O, N=N and to other second periodic elements, e.g. C=O, C=N, N=N.

This is a Short Answer Type Questions as classified in NCERT Exemplar

In the p block, some elements are metallic some elements are non-metallic while some elements are metalloids in nature. The oxides of metals are basic in nature and that of oxides of nonmetals are acidic in nature.

Acidic oxide SO₂ + H₂O →H₂SO₃

Basic oxide  Tl₂O + 2HCl → 2TlCl +H₂O

Amphoteric oxide

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Al₂O₃ +2 NaOH → 2NaAlO₂ + H₂O

Reaction with water

B₃O₂ +3H₂O → 2H₃BO₃

P₄O₁₁  + 6H2O  → 4H₄PO₃

Cl₂ O₇ + H2O → HClO₄

This is a Short Answer Type Questions as classified in NCERT Exemplar

The electronic configuration of Na is [Ne]3s¹ while that of Mg is [Ne]3s² as Mg possesses a fully filled 3s orbital thus its first ionisation enthalpy is higher than that of Na.

After losing one electron Na obtained the electronic configuration of Ne while Mg acquired the electronic configuration of Na thus the second ionisation energy of Na is higher than that of Mg.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Exothermic reaction- The reaction in which heat is released

For example- CaO + CO₂ → CaCO₃ + Heat

Endothermic reaction- The reaction in which heat is absorbed.

For example- 2NH₃ + heat → N₂ +3H₂

 This is a Short Answer Type Questions as classified in NCERT Exemplar

(i) S

The ionization enthalpy increases across the period and decreases down the group. N possesses a half filled 2p orbital which provides it extra stability due to symmetry.

(ii)P

The non-metallic character increases across the period

The deviation in the ionization enthalpy of some elements from the general trend can be explained by the points as given below:-

(i) The fully filled and half filled orbital provide extra stability due to the symmetry

(ii) The effective nuclear charge

(iii) The e^- - e^- repulsion which lead to instability

This is a Short Answer Type Questions as classified in NCERT Exemplar

(a) As the effective nuclear charge increases and shielding effect decreases across the periods thus the electronegativity increases on moving from left to right in the periodic table.

(b) As the number of shells increases down the group thus its ionisation enthalpy decreases down the group.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Metallic character- Tendency to lose electrons.

Non-Metallic character- Tendency to accept electrons.

Across the period the metallic character decreases whereas non-metallic character increases across the period.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The radius of Na⁺ ion is smaller than that of Na is due to the following reasons:-

(i) The effective nuclear charge of Na⁺

(ii) The disappearance of 3s orbital from its outermost shell electronic configuration.

This is a Short Answer Type Questions as classified in NCERT Exemplar

The electronegativity of the alkali metals decreases down the group due to the increase of the shells thus Cs (Caesium) is the least electronegative metal in the alkali metals.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii)

All the given ions are isoelectronic species thus their radii depend upon the charge more the negative charge higher would be the atomic radii and higher the positive charge lesser would be the atomic radii.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv)

Actinoids possess Z=90-103

Terbium is lanthanoid.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

The orbitals nearer to the nucleus possess more shielding effect.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

As Mg contains fully filled 3s orbital thus its ionisation is higher than that of Na and Al.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Gadolinium is a lanthanoid which belongs to the f-block elements with electronic configuration (n-2)f¹⁴ (n-1)d⁰-¹ns²

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

As per Aufbau principle 3d orbitals are filled first than 4s as due to lower n value.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

The electron gain enthalpy decreases down the group, however as the size of F atom is much smaller thus it leads to repulsion upon addition of electrons on its 2p orbital for this electron gain enthalpy of Cl is higher than that of F.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Period number indicates the maximum value of principal quantum number

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

The lanthanoids which belong to the f-block elements are the elements where 4forbitals are filled progressively.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv)

I⁺ possess the highest effective nuclear charge whereas I^- possess the least among the given species.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

High amount of energy has to be supplied in order to overcome the e^- - e ^- repulsion that arises when O^- gets converted to O^2- by accepting an electron

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii) d-block

Lanthanum is the element with the atomic number 57 and it belongs to the d-block.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

The electron gain enthalpy of 2p orbital is higher than that of the 3s orbital

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii) & (iii)

Both P and S have vacant d-orbital for which they execute extended covalency.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) & (iii)

Alkali metals and alkaline earth metals have low ionization enthalpy.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option   (i)& (iv)

The s-block elements and p-block elements together are called representative elements.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iv)

Both S and Cl have the higher tendency to gain electrons in order to attain stable the nearest noble gas configuration i.e. of Argon.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i), (iii)& (iv)

He with 1s² electronic configuration has the highest electron gain enthalpy in the periodic table. In any period the alkali metals have the lowest effective nuclear charge in that particular period for which its atomic radius is the highest in that period.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (ii)& (iii)

Isoelectronic species/ions are those species that possess the same number of electrons.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii)& (iii)

The ionization enthalpy of N is higher than that of F because it possess half filled orbital which provide it extra stability due to symmetry.

 As the atomic size of F is much smaller thus its electron gain enthalpy is lower than that of the Cl

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iv)

Both electronegativity and metallic character do not possess units as they both are qualitative properties not a quantitative property.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iii)

Ionic radii decreases with the increase of the effective nuclear charge and increases with the increase of shielding effect or e^- - e ^- repulsion as it outweighs the effective nuclear charge effect.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iii)

Aluminium is the element which is a metal and good conductor of electricity.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (ii)

The ionization energy depends upon two factors

(a) The effective nuclear charge.

(b)The e^- -e ^– repulsions.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (iii)

The Be has the electronic configuration of [He]2s² while B has the electronic configuration of [He]2s²2p¹ . As in Be the outermost orbital i.e. 2s is fully filled thus it provides extra stability to the Be compared to B. Apart from this 2s electron is more penetrated compared to the 2p electron for which 2p electron faces more shielding effect than that of 2s electron.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Option (ii)

The electron gain enthalpy decreases down the group due to the increase in the atomic size which leads to decrease in the effective nuclear charge.

This is a Matching Type Questions as classified in NCERT Exemplar

Be                                                  111

C                                                    77

O                                                    66

B                                                     88

N                                                    74

This is a Matching Type Questions as classified in NCERT Exemplar

i )                            b  1681    3374    – 328

ii )                           a   419     3051     – 48

iii)                           d   2372   5251    + 48

iv)                           c   738     1451    – 40

This is a Matching Type Questions as classified in NCERT Exemplar

i                                           d          + 48

ii                                          a          – 53

iii                                         b          – 328

iv                                         c          – 141

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 × no. of moles of H₂O = $\left(\frac{270}{18}\times 2\right)$ moles

              Mass of C = $\frac{330}{44}\times 12$ gm = 90 gm

              Mass of H = $\frac{270}{18}\times 2\times 1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}\times 100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}\times 100\mathrm{\%}=25\mathrm{\%}\end{array}$  

Energy of 1 photon =  $\frac{hc}{\lambda }$

Energy of 1 mole of photon = $N_A\times \frac{hc}{\lambda }$ ${}_{}\frac{}{}$

$=\frac{6.022\times 10^{23}\times 6.63\times 10^{-34}\times 3\times 10^8}{300\times 10^{-9}}$

= 0.399 × 106 J = 399 KJ

Bond order of   $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$

Bond order of $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$

Bond order of   $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

Metal                                 Ag                        Ga                        Cs                         Hg

              Melting point                   961°C                  29.7°C                 28.5°C                 -38.3°C

In Hall Heroult process electrolysis is carried out using carbon anode. The liberated

O2 at anode reacts with anode.

Cathode : $A{l}^{3+}\left(melt\right)+3ek\to A\mathcal{l}\left(\mathcal{l}\right)$

Anode : $\begin{array}{l}C\left(s\right)+O^{2-}\left(Melt\right)\to CO\left(g\right)+2e\\ C\left(s\right)+2O^{2-}\left(Melt\right)\to C{O}_2\left(g\right)+4e\end{array}$

Net Reaction 2Al2O3 + 3C $4A\mathcal{l}+3C{O}_2$

NaOH is obtained by electrolysis of NaCl. Solution of NaCl using Hg cathode

$\begin{array}{l}Cathode:N{a}^{+}+e\underset{}{\overset{Hg}{\to }}Na-Amalgam\\ Anode:2C{l}^{-}\to C{l}_2+2e\\ 2Na-Amalgam+2H_{2}O\to 2NaOH+2Hg+H_2\end{array}$

Fire extinguisher contains sodium Bicarbonate (NaHCO3), also known as baking soda.

Phosphorous can expand its covalency beyond 4, as it has vacant 3d orbitals. Nitrogen can't expand its covelency beyond 4, as it does not posses vacant 3d- orbitals.

The value of crystal field spilitting energy ${}_{}$ $\left({\Delta }_0\right)$ increases down the group. 5d series member will have more ${}_{}$ ${\Delta }_0$ compared to 3d series or 4d series member.

CH₄, O₃ & H₂O are green-house gases.

 Bond is weaker then $$ - $\sigma$ bond.

For the reaction C? H? → C? H? + H? , calculate the enthalpy change (ΔH).

ΔH = [Bond energy (C-C) + 6 × Bond energy (C-H)] - [Bond energy (C=C) + 4 × Bond energy (C-H) + Bond energy (H-H)]

ΔH = 347 + 2 (414) - 611 - 436 = 128 kJ/mol.

For an acidic buffer solution, pH = pKa + log ( [Base]/ [Acid]).

Given pH = 5.74 and pKa = 4.74.

5.74 = 4.74 + log ( [Base]/1).

1 = log ( [Base]).

[Base] = 10M.

AX is a diatomic molecule with a bond order of 2.5.

The compound is NO. The total number of electrons = 15 (7+8).

Please consider the following Image

3-Hydroxy propanal

If 7.8g of C? H? O (molar mass 56 g/mol ) is formed, calculate the initial weight of 3-hydroxy propanal (molar mass 74 g/mol ).

Weight = (7.8/56) × 74 × (100/64) [Assuming 64% yield, though the number seems out of place].

Ans ≈ 16 g.

Please consider the following Image

Given K_f = 1.85 K kg mol? ¹ for a solution with molality of 2 m.

ΔT_f = I * K_f * m

3.885 = I * 1.85 * 2

The van't Hoff factor, I = 1.05.

i = 1 + (n-1)α. For an electrolyte dissociating into 2 ions, n=2.

1.05 = 1 + (2-1)α.

The degree of dissociation, α = 0.05 or 50 × 10? ³.

Combustion of ethane: C? H? + 7/2 O? → 2CO? + 3H? O.

Moles of C? H? = 3g / 30 g/mol = 0.1 mol.

Moles of H? O produced = 0.3 mol.

Number of H? O molecules = 0.3 × 6.023 × 10²³ ≈ 18.06 × 10²².

So, x = 18.

For the complex K? [Cr (oxalate)? ], the central metal ion is Cr³?

Electronic configuration of Cr (24) is [Ar] 4s¹3d?

Electronic configuration of Cr³? is [Ar] 4s?3d³.

The number of unpaired electrons in Cr³? is 3.