Chemistry 3 Marks Question

1. 1 mol of NH₃ = 17g NH₃ = 6.022 × 10²³ molecules of NH₃ 

1 atom of NH₃ contains = 7 + 3 = 10 protons

$\therefore$ The number of protons in 1 mol of NH₃ = 6.022 × 10²⁴ protons.

Number of protons in 34mg of NH₃ $=\frac{(6.022\times10^{24}\times34)}{17\times1000}=1.2044\times10^{22}\text{ protons.}$

2. Mass of one proton = 1.6726 × 10^-27kg

$\therefore$ Mass of 1.2044 × 10²² protons = (1.6726 × 10^-27) × (1.2044×10²²)kg = 2.0145 × 10^-5kg.

No, there will be no effect of temperature and pressure.

1. Number of atoms in ¹⁴C in 1 mole = 6.022 × 10²³ atoms

1 atom of ¹⁴C contains = 14 - 6 = 8 neutrons.

$\therefore$ The number of neutrons in 14g of ¹⁴C = 6.022 × 10²³ × 8 neutrons

Number of neutrons in 7mg $=\frac{(6.022\times10^{23}\times8\times7)}{14000}=2.4088\times10^{21}\text{ netrons}$

2. Mass of one neutron = 1.674 × 10^-27kg

Mass of total neutrons in 7g of ¹⁴C = (2.4088 × 10²¹)(1.675 × 10^-27kg) = 4.035 × 10^-6kg

1. [He] 2s¹

The electronic configuration of the element is [He] 2s¹ = 1s² 2s¹.

$\therefore$ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

2. [Ne] 3s² 3p³

The electronic configuration of the element is [Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³.

$\therefore$ Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

3. [Ar] 4s² 3d¹

The electronic configuration of the element is [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.

$\therefore$ Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s² 3d¹ is scandium (Sc).

1. s-Orbital is spherically symmetrical, i.e. it is non directional.
2. Transition is from n₁ = 5 to n₂ = 2 state,

$\Delta\text{E}=2.18\times10^{-18}\text{J}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$

$=2.18\times10^{-18}\text{J}\Big(\frac{1}{5 ^2}-\frac{1}{3^2}\Big)$

$=-4.58\times10^{-19}\text{J}$

The negative sign of energy means it is an emission energy.

Frequency $=\frac{\text{Energy}}{\text{Plant's constant}}$

$\Rightarrow\text{v}=\frac{\Delta\text{E}}{\text{h}}=\frac{4.58\times10^{19}\text{J}}{6.626\times10^{-34}\text{Js}}$

$=6.91\times10^{14}\text{Hz}$

Wavelength $=\frac{\text{Velocity}}{\text{Frequency}}$

$\Rightarrow\lambda=\frac{\text{c}}{\text{v}}=\frac{3\times10^8\text{ms}^{-1}}{6.91\times10^{14}\text{Hz}}$

$\lambda=0.434\times10^{-6}\text{m}$

$\lambda=434\times10^{-9}\text{m}=434\text{nm}$ $[\because1\text{nm}=10^{-9}\text{m}]$

1. Given, $\text{n}=5.09\times10^{14}\text{s}^{-1},$

$\text{c}=3\times10^8\text{ms}^{-1},\lambda=\ ?$

Wavelenth, $\lambda=\frac{\text{c}}{\text{v}}=\frac{3\times10^8\text{ms}^{-1}}{5.09\times10^{14}\text{s}^{-1}}$

$=5.89\times10^{-7}\text{m}$

$=5.89\times10^{-7}\times10^9\text{nm}=589\text{nm}$

2. (a) 3p_y; (b) 1s
3. Spin quantum number, Pauli exclusion principle.

1. $\text{m}=9.1\times10^{-28}\text{g}=9.1\times10^{-31}\text{kg}$

$\text{K.E}=3.0\times10^{-25}\text{J}$

$\frac{1}{2}\text{mv}^2=3.0\times10^{-25}\text{J}$

$\text{K.E}=\frac{1}{2}\text{mv}^2$

$\Rightarrow\text{V}=\sqrt{\frac{2\text{KE}}{\text{m}}}$

$\lambda=\frac{\text{h}}{\text{mV}}=\frac{\text{h}}{\text{m}\times\sqrt{\frac{2\text{K.E}}{\text{m}}}}=\frac{\text{h}}{\sqrt{2\times\text{K.E}\times\text{m}}}$

$\lambda=\frac{\text{h}}{\sqrt{2\text{m}\times\text{K.E}}}=\frac{6.6\times10^{-34}\text{Js}}{\sqrt{2\times9.1\times10^{-31}\times3\times10^{-25}}}$

$\lambda=\frac{6.6\times10^{-34}}{\sqrt{54.6\times10^{-56}}}=\frac{6.6\times10^{-34}}{7.39\times10^{-28}}$

$=8.93\times10^{-7}\text{m}=893\text{nm}=8930\mathring{\text{A}}$

2. When a beam of light having frequency more than threshold frequency is made to fall on metals like alkali metals, electrons are ejected. These electrons are called photoelectrons and this phenomenon is called photoelectric effect.

1. Differences between orbit and orbitals:

2. Uncertainty in speed, $\Delta\text{V}=\frac{0.005}{100}\times600\text{m/ s}$

$=0.03\text{ms}^{-1}$

Heisenberg Uncertainty Principle,

$\Delta\text{x} \times\text{m}\Delta\text{V}=\frac{\text{h}}{4\pi},\Delta\text{x}=$ Uncertainty in position,

$\Rightarrow\Delta\text{x}=\frac{6.626\times10^{-34}\text{Js}}{4\times\frac{22}{7}\times9.11\times10^{-31}\text{kg}\times0.03\text{ms}^{-1}}$

$=1.93\times10^{-3}\text{m}$

1. $\text{m}=9.1\times10^{-31}\text{kg},$

$\text{c} =3.0\times10^8\text{ms}^{-1}$

$\lambda=\ ?,\text{h}=6.626\times10^{-34}\text{Js}$ or $\text{Kg}/\ \text{m}^2\text{s}^{-1}$

Applying de Broglie equation:

$\lambda=\frac{\text{h}}{\text{mc}}=\frac{6.626\times10^{-34}\text{kg/ m}^2\text{s}^{-1}}{9.1\times10^{-31}\text{kg}\times3.0\times10^8\text{ms}^{-1}}$

$\lambda=\frac{6.626}{27.3}\times10^{-34+31-8}\text{m}$

$\lambda=\frac{6.626}{27.3}\times10^{-11}=\frac{66.26}{27.3}\times10^{-12}\text{m}$

$\lambda=2.427\times10^{-12}\text{m}$

2. Angular momentum mvr $=\frac{\text{nh}}{2\pi}$

For 5^th energy level, $\text{n}=5$

$\therefore\text{mvr}=\frac{5\text{h}}{2\pi}$

1. For $\text{n}=2,$

$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2},$

$\text{l}=1,\text{m}_\text{l}=1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}.$

Therefore, a total of 2 + 6 = 8 electrons are present.

2. For $\text{n}=3,$ when

$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=\frac{1}{2},-\frac{1}{2}$

$\text{l}=2,\text{m}_\text{l}=-2,-1,0,+1,+2,,\text{m}_\text{s}$

$=+\frac{1}{2},-\frac{1}{2}$

Therefore, a total of 2 + 6 + 10 = 18 electlons aie present.

3. For $\text{n}=4,$ when

$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=3,\text{m}_\text{l}=-2,-1,0+1,+2,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=3,\text{m}_\text{l}=-3,-2,-1,0,+1,+2,+3,\text{m}_\text{s}$

$=+\frac{1}{2},-\frac{1}{2}$

Therefore, a total of 2 + 6 + 10 + 14 = 32 electrons are present.

1. For He⁺ ion, Z = 2, third orbit, n = 3

$\text{r}_3(\text{He}^+)=\frac{3^2\text{h}^2}{4\pi^2\text{m}\times2\times\text{e}^2}$

$=\frac{9}{2}\Big[\frac{\text{h}^2}{4\pi^2\text{me}^2}\Big]=\frac{9}{2}\times0.529=2.380\mathring{\text{A}}$

2. For Li²⁺ ion, Z = 3, second orbit, n = 2

$\text{r}_2(\text{Li}^{2+})=\frac{2^2\text{h}^2}{4\pi^2\text{m}\times3\times\text{e}^2}=\frac{4}{3}\Big[\frac{\text{h}^2}{4\pi^2\text{me}^2}\Big]$

$=\frac{4}{3}\times0.529=0.7053\mathring{\text{A}}$

1. $1\text{s}^2\ 2\text{s}^2,2\text{p}^2_\text{x},2\text{p}^2_\text{y},2\text{p}^2_\text{z},3\text{s}^2$

2. $1\text{s}^2\ 2\text{s}^2,2\text{p}^1_\text{x},2\text{p}^1_\text{y},2\text{p}^1_\text{y},2\text{p}^1_\text{z}$

3. $1\text{s}^2\ 2\text{s}^2,2\text{p}^6,3\text{s}^2,3 \text{p}^6,4\text{s}^2,3\text{d}^2$

4. $1\text{s}^2\ 2\text{s}^2,2\text{p}^6,3\text{s} ^2,3\text{p}^6,3\text{d}^5,4\text{s}^1$

Isoelectronic are the species having same number of electrons.

A positive charge means the shortage of an electron.

A negative charge means gain of electron.

Number of electrons in Na⁺ = 11 - 1 = 10

Number of electrons in K⁺ = 19 - 1 = 18

Number of electrons in Mg²⁺ = 12 - 2 = 10

Number of electrons in Ca²⁺ = 20 - 2 = 18

Number of electrons in S^2- = 16 + 2 = 18

Number of electrons in Ar = 18

Hence, the following are isoelectronic species:

1. Na⁺ and Mg²⁺ (10 electrons each)
2. K⁺, Ca²⁺, S^2- and Ar (18 electrons each)

$\lambda=\frac{6.626\times10^{-34}\text{Js}}{(9.10939\times10^{-31}\text{kg})(811.579\text{ ms}^{-1})}$

$\lambda=8.9625\times10^{-7}\text{m}$

Hence, the wavelength of the electron is 8.9625 × 10^–7m.

1. H^– ion

The electronic configuration of H atom is 1s¹.

A negative charge on the species indicates the gain of an electron by it.

$\therefore$ Electronic configuration of H^– = 1s²

2. Na⁺ ion

The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.

A positive charge on the species indicates the loss of an electron by it.

$\therefore$ Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s⁰ or 1s² 2s² 2p⁶

3. O^2– ion

The electronic configuration of 0 atom is 1s² 2s² 2p⁴.

A dinegative charge on the species indicates that two electrons are gained by it.

$\therefore$ Electronic configuration of O^2– ion = 1s² 2s² p⁶

4. F^– ion

The electronic configuration of F atom is 1s² 2s² 2p⁵.

A negative charge on the species indicates the gain of an electron by it.

$\therefore$ Electron configuration of F^– ion = 1s² 2s² 2p⁶

1. $\text{Radius of zinc atom }=\frac{\text{Diameter}}{2}$

$=\frac{2.6\mathring{\text{A}}}{2}$

$=1.3\times10^{-10}\text{m}$

$=130\times10^{-12}\text{m}=130\text{pm}$

2. Length of the arrangement = 1.6cm

= 1.6 × 10^–2m

Diameter of zinc atom = 2.6 × 10^–10m

$\therefore$ Number of zinc atoms present in the arrangement

$=\frac{1.6\times10^{-2}\text{m}}{2.6\times10^{-10}\text{m}}$

$=0.6153\times10^8\text{m}$

$=6.153\times10^7$