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Covalent compounds have strong forces of attraction within the molecule, but the inter-molecular forces are very small due to which they are volatile. This results in the low melting and boiling point of compounds making them volatile.

(a) Element E: Carbon

(b) Allotrope A: Buckminsterfullerene

(c) Allotrope B: Graphite

(d) Allotrope C: Diamond

(e) C

(f) B

The allotropic forms is buckminsterfullerene is-

Carbon

The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy. The allotropic forms of carbon are graphite, diamond, etc.

Covalent Bond

A covalent bond is defined 'as the force of attraction arising due to mutual sharing of electrons between the two atoms.' The combining atoms may share one, two or three pairs of electrons. The covalent bond is formed between two similar or dissimilar atoms by a mutual sharing of electrons, which are counted towards the stability of both the participating atoms. When the two atoms combine by mutual sharing of electrons, each of the atoms does so, in order to acquire stable configuration of the nearest noble gas. A small line (-) between the two atoms is represents a covalent bond. The compounds formed due to covalent bonding are called covalent compounds.

(i) Ethanoic acid provides an acidic environment where bacteria cannot survive, preserving the pickles for a longer time. 

(ii) Both diamond and graphite are allotropes of the same element carbon. The only difference is in their physical properties due to the difference in their structure. The chemical properties remain the same. Carbon when burnt in the presence of oxygen gives carbon dioxide (CO₂). Since, diamond and graphite are chemically carbon, both release CO₂ upon burning.

Properties of Covalent Compounds

• The covalent compounds do not exist as ions but they exist as molecules

• They exist at room temperature, as liquids or gases. However, a few compounds also exist in the solid state e.g. urea, sugar, etc.

• The melting and boiling points of covalent compounds are generally low

• Covalent compounds are generally insoluble or less soluble in water and in other polar solvents

• These are poor conductors of electricity in the fused or dissolved state

• Since the covalent bond is localized in between the nuclei of atoms, it is directional in nature

• A covalent bond can be formed in different ways. When a bond is formed by mutual sharing of one pair of electrons it is known as a 'single covalent bond', or simply 'a single bond'. When a bond is developed due to mutual sharing of more than one pairs of electrons it is termed as 'multiple covalent bond'. Such bonds can be a double covalent bond or a triple covalent bond.

X is ethanoic acid (CH₃COOH). It is acidic and hence turns blue litmus red. It releases carbon dioxide (CO₂) when treated with Na₂CO₃ and NaHCO₃ which is responsible for effervescence. 

2CH₃COOH + Na₂CO₃→ 2CH₃COONa + CO₂↑ + H₂O 

CH₃COOH + NaHCO₃→ CH₃COONa + CO₂↑ + H₂O

CH₃COCH₃: Propanone 

C₂H₅COOH : Propanoic acid

(i) Butyne 

(ii) Ethanol 

(iii) Propanone

Disadvantages of incomplete combustion:

(i) It leads to the formation of soot which is nothing but unburnt carbon which pollutes the atmosphere, blackens cooking utensils.

(ii) It leads to the formation of an extremely poisonous gas called carbon monoxide.

(i) C₂H₄O₄: CH₃-COOH - ethanoic acid

(ii) C₂H₅OH : ethanol

(iii) X is CH₃COOC₂H₅ - ethyl acetate [ethyl ethanoate-ester]

CH₃CH₂OH – Ethanol 

Functional group: alcohol (–OH)

2CH₃COOH+Na₂CO₃ →2CH₃COONa+H₂O+CO₂↑

(a) Air holes of a gas burner have to be adjusted b ecause blackening of vessels show that the air holes of the gas stove are getting blocked and hence the fuel is not burning completely (due to insufficient supply of oxygen).

(b) Some of the detergents (synthetic) are not bio-degradable, that is they cannot be decomposed by micro organisms like bacteria and hence cause water pollution.

Ethanol burns in air to produce carbon dioxide and water. 

C₂H₅OH + 3O₂ ⎯⎯→ 3CO₂ + 3H₂O + heat

Carbon forms large number of compounds due to its property of catenation, i.e. self linking. They form isomeric compounds i.e. compounds with same molecular formula but different structural formula.

Two forms of energy obtained are heat energy and light energy.

(a) Propanol

(b) C₃H₇OH

(c) Propane

(d) -OH (alcohol)

(a) Graphite

(b) Graphite

(c) Hydrocarbons

(d) Alkene

(e) Alkynes

(f) Unsaturated

(g) Two

(h) Low; covalent

(i) Catenation

(j) Alkenes

(k) Ethene

(l) Ethyne

(a) Alkynes, C_nH_2n-2

First member: Ethyne

(b) Second member: C₃H₆

Fourth member: C₅H₁₀

(c) Third member: C₃H₈

Fifth member: C₅H₁₂

The first member of ketone - Propanone.

(a) Hydrocarbons

(b) Carbon forms covalent bonds because it can achieve the inert gas electron arrangement only by sharing of electrons.

Carbon can form bond by sharing of its electrons with the electrons of other carbon atom or with other element and attain noble gas configuration.

(a)The atomic number of carbon is 6. Its electronic configuration is 2,4.

(b)Carbon forms covalent bonds because it can achieve the inert gas electron arrangement only by sharing of electrons.

(c)Diamond, graphite and buckminsterfullerene.

Energy of electron in nth orbit of H atom is given by

= {-2.18 x 10^-11 ergs}/{n²}

Thus, E₁ = -2.18 x 10^-11 ergs

= -218 x 10^-13 ergs

and E_s = {-2.18 x 10^-11}/{(5)²} ergs

= -8.72 x 10^-13 ergs

Energy required for the electronic transition

ΔE = E₅ - E₁

= -8.72 x 10^-13 ergs - (-218 x 10^-13 erg)

= 2.08 x 10^-11 ergs

Wavelength of the emitted radiation,

λ = ch/ΔE = {3 x 10¹⁰ cm x 6.626 x 10^-27 ergs}/{2.09 x 10^-11 ergs}

or, λ = 9.511 x 10^-6 cm

= 951.1 x 10^-8 cm = 956 Å

B) Li, Be, C, N

\(\Delta E\) = \(E_4-E_1\) = 2.18 x 10^-11\(\Big(\frac{1}{n_1^2}-\frac{1}{n_2^2}\Big)\) erg

= 2.18 × 10^-11\(\Big(\frac{1}{(1)^2}-\frac{1}{(4)^2}\Big)\)

= 2.18 × 10^-11x \(\frac{16-1}{16}\)

= 2.18 × 10⁻¹¹ x \(\frac{15}{16}\)

= 2.04375 × 10⁻¹¹ erg

\(\therefore\) 1 erg = 10⁻⁷ J

\(\therefore\) \(\Delta E\) = 2.04375 × 10⁻¹¹ × 10⁻⁷ J

= 2.04375 × 10⁻¹⁸ J

\(\Delta E\) = \(\frac{hc}{\lambda}\)

Or \(\lambda = \frac{hc}{\Delta E}\)

= \(\frac{6.626\times10^{-34}Js\times3\times10^8m\,sec^{-1}}{2.04375\times10^{_18}J}\)

= 9.726 × 10⁻⁸ m

= 972.6 × 10⁻¹⁰ m

= 972.6 Å

\(\Delta\)E = 2.18 x 10^-18\(\Big[\frac{1}{n_1^2}-\frac{1}{n_2^2}\Big]\)

Given n₁ = 1 and n₂ = 4

\(\Delta\)E = 2.18 x 10^-18\(\Big[\frac{1}{(1)^2}-\frac{1}{(4)^2}\Big]\)

= 2.18 x 10^-18\(\Big[\frac{1}{1}-\frac{1}{16}\Big]\)

= 2.18 x 10^-18\(\Big[\frac{16-1}{16}\Big]\)

= 2.18 x 10^-18\(\big(\frac{15}{16}\big)\)J

= 2.18 x 10^-18 J

\(\therefore\) \(\Delta\)E = hv

v = \(\frac{\Delta E}{h}\)

=  \(\frac{2.04\times10^{-18}}{6.626\times10^{-34}j\,sec}\)

= 3.08 × 10¹⁵ Hz

\(\lambda=\frac{c}{v}=\frac{3\times10^8ms^{-1}}{3.08\times10^{15}s^{-1}}\)

= 9.7 × 10⁻⁸ m

= 97 × 10⁻⁹ m

= 97 nm

Energy E = -13.6 x \(\frac{z^2}{n^2}\)eV

For He⁺ z = 2

Value of n for first orbit = 1

\(\therefore\) E = -13.6 x \(\frac{(2)^2}{(1)^2}\)

= -54.4 eV

Radius of this orbit,

r_n = \(\frac{52.9(n^2)}{z}\)

r₁ = \(\frac{52.9\times1^2}{2}\)

= 26.45 pm

= 0.2645 Å

Radius of n^th Bohr orbit, r_n = \(\frac{n^2b^2}{4 \pi^2 m Ze^2}\)

For hydrogen atom Z = 1, first orbit n = 1

r₁ = \(\frac{b^2}{4 \pi^2 m e^2}\) = 0.529 Å

(i) For He⁺ ion, Z = 2, third orbit, n = 3

r₃(He⁺) = \(\frac{3^2b^2}{4 \pi^2m \times 2 \times e^2}\)

= \(\frac{9}{2}\Big [\frac{b^2}{4 \pi^2me^2} \Big]\) = \(\frac{9}{2} \)x 0.529 = 2.380 Å

(ii) For Li²⁺ ion, Z = 3, Second orbit, n = 2

r₂(Li²⁺) = \(\frac{2^2b^2}{4 \pi^2 m \times 3 \times e^2}\) = \(\frac{4}{3} \Big[ \frac{b^2}{4 \pi^2 me^2} \Big]\)

= \(\frac{4}{3}\)x 0.529 = 0.7053 Å

(i) E = -2.18 x 10^-18 x \(\frac{z^2}{n^2}J\) atom^-1

For H − atom, z = 1

value of n for fourth orbit = 4

\(\therefore\) E = -2.18 x 10^-18\(\frac{(1)^2}{(4)^2}\)

= \(\frac{-2.18\times10^{-18}}{16}\)

= -0.136 x 10^-18

=-1.36 x 10^-19 J

(ii) Radius of third orbit for H-atom

r_n = \(\frac{52.9n^2}{z}\)

r₃ = \(\frac{52.9\times(3)^2}{1}\)

= 52.9 × 9

= 476.1 pm

Q Radius of n^th Bohr orbit

r_n = \(\frac{0.529n^2}{z}\)

For H-atom, z = 1,

For first orbit n = 1

(i) For He⁺ ion, z = 2, third orbit, n = 3

r₃ = \(\frac{0.529\times(3)^2}{2}\)

= \(\frac{0.529\times9}{2}\)

= 2.380 Å

(ii) For Li²⁺ion, z = 3

Second orbit, n = 2

\(\therefore\) r₂ = \(\frac{0.529\times(2)^2}{3}\)

= \(\frac{0.529\times4}{3}\)

= 0.7053 Å

(i) Energy (E) = \(\frac{hv}{\lambda}\); h = 6.6 × 10⁻³⁴Js

C = 3 x 10⁸ ms^-1

E = \(\frac{6.6\times10^{-34}Js\times3\times10^8m}{4\times10^{-7}m}\)

= 4.97 × 10⁻¹⁹ J

= \(\frac{4.97\times10^{-19}J}{1.602\times10^{-9}}\)

= 3.10 eV.

Energy of the photon in eV = 3.10

(ii) Kinetic energy of the emission of photo-electrons (K.E)

= hv − \(\omega_o\)(where \(\omega_o\) = work function)

= 3.10 eV − 2.13 eV = 0.97 eV.

T = 2.0 x 10^-10 s

Frequency, v = 1/T = {1}/{2.0 x 10^-10} = 5 x 10⁹ s^-1

v = vλ

i.e., 3 x 10⁸ = 5 x 10⁹ x λ

λ = {3 x 10⁸}/{5 x 10⁹} = 6.0 x 10^-2 m

Wave number, bar v = 1/λ = {1}/{6.0 x 10^-2} = 100/6 = 16.66 m^-1

Wavelength of light, λ = 4000 pm = 4000 x 10^-12 m

Energy provided, E = 1 J

Energy per photon = hc/λ

If N is the number of photons that provide 1 J energy, we can write

1 J = N x hc/λ

or, N = {1 J x λ}/{hc}

= {15 x 4000 x 10^-12 m}/{6.626 x 10^-34 Js x 3 x 10⁸ m/s}

= 2.012 x 10¹⁶

For N-shell principal quantum number n = 4 

∴ Total number of orbitals in N-shell = n² = 4² = 16. 

The total number of subshells in N-shell = n 

= 4.

The four subshells with their azimuthal quantum numbers and the constituent number of orbitals are as shown below :

5th energy shell.

It measures the electron probability density at a point in an atom.

Since, Cs is the most electropositive element of all so it has the minimum ionization energy and contains the maximum capacity to lose electrons. That's why Cs shows maximum photoelectric effect.

The photoelectric effect is maximum in Cs as it has the lowest first ionization energy among alkali metals.
On moving down the group of alkali metals, the ionization energy decreases and the removal of the electron becomes easy. Hence Cs has the maximum photoelectric effect

C) 400 nm – 700 nm

400nm to 750nm

Correct option is  B) 2

It is because they are associated with fixed amount of energy.

Lyman series of lines of the hydrogen spectrum lies in the ultraviolet region.

D) Heavier atoms

1. S: 1s2 2s2 2p6 3s2 3p4 

2.  K: 1s2 2s2 2p6 3s2 3p6 4s1 

3. Ti: 1s2 2s2 2p6 3s2 3p6 3d2 4s2 

4. Sn: 1s2 2s2 2p6 3s2 3p6 3d104s2 4p6 4d105s2 5p2 

Large number of lines are there in hydrogen spectrum because the gas contains large number of atoms and therefore large number of electrons and for a particular line to form there are large number of electrons falling from one orbit to another orbit.

Correct option is A) shells