Structure of AtomNEET Chemistry · Class 11 · NCERT Chapter 2

Solution

For $n=3$ and $l=1$, the subshell is 3p. The magnetic quantum number $m_l=0$ specifies one specific orbital in the 3p subshell, so the maximum number of orbitals is 1. Option (a) is correct.

Solution

The energy of a photon is given by $E=\frac{hc}{\lambda }$. Substituting the values, $E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{45\times 10^{-9}}=4.42\times 10^{-18}\text{J}$, so option (d) is correct.

Solution

The correct order of ionic radii is ${\text{O}}^{2-}>{\text{F}}^{-}>{\text{Na}}^{+}$. This is because ${\text{O}}^{2-}$ has the most electrons and the largest ionic radius, followed by ${\text{F}}^{-}$, and then ${\text{Na}}^{+}$, which has the fewest electrons and the smallest ionic radius. Option (b) is correct.

Solution

${\text{Be}}^{2+}$ has 2 electrons, and ${\text{Li}}^{+}$ also has 2 electrons, making them isoelectronic. NCERT XI chapter Structure of Atom defines isoelectronic species as those with the same number of electrons, so option (b) is correct.

Solution

The formation of ${\text{O}}^{2-}$ is unfavourable because the second electron addition results in significant electron repulsion, which outweighs the stability gained by achieving the noble gas configuration of neon. This is consistent with the endothermic nature of the second electron affinity, so option (b) is correct.

Solution

Ti(22)=1s22s22p63s23p64s23d2 Order of energy is 3s 3p 4s 3d

Solution

When Agrobacterium tumifaciens infects the host plant, it will transfer a part of DNA called t-DNA without any human interference so called natural genetic engineer.

Solution

1 λ = R =(1 h1 2 − 1 h2 2) Wavelength = 1 λ = R[ 1 22] = R 4 = 107 4 = 0.25× 107 m−1

Solution

Electron occupying same orbital have different spin quantum number.

Solution

𝐵𝑒2: σ 1𝑠2< σ∗1𝑠2< σ2𝑠2< σ∗2𝑠2 𝑂2: σ1𝑠2< σ∗1𝑠2< σ2𝑠2< σ∗2𝑠2< σ 2𝑝𝑧2 <[π2𝑝𝑦1+1= π2𝑝𝑦1+1]<[π∗2𝑝𝑥1= π∗2𝑝𝑦1] 𝑁2: σ1𝑠2< σ∗1𝑠2< σ2𝑠2< σ∗2𝑠2<[π2𝑝𝑥1+1= π2𝑝𝑦1+1]< σ 2𝑝𝑧2 𝐶2: σ1𝑠2< σ∗1𝑠2< σ2𝑠2< σ∗2𝑠2<[π2𝑝𝑥1+1= π2𝑝𝑦1+1]

Solution

N2(g) + 3H2(g) → 2NH3(g) To produce 20 moles of ammonia, we need 30 mole of H2.

Solution

Basic character in aq. Medium = 2 ° > 1 ° > 3 ° amines

Solution

The Bohr model is valid for hydrogen-like atoms, which have one electron. Hydrogen, singly ionised helium, and deuteron all fit this criterion. Singly ionised neon, however, has more than one electron, making the Bohr model inapplicable. Thus, option (d) is correct.

Solution

The atomic number (71) gives the number of protons and electrons, and the mass number (175) minus the atomic number gives the number of neutrons: $175-71=104$. Therefore, the correct option is (a).

Solution

The spin-only magnetic moment $\mu$ is given by $\mu =\sqrt{n(n+2)}\text{BM}$, where $n$ is the number of unpaired electrons. For ${\text{Cr}}^{2+}$, the electronic configuration is $3d^4$, with 4 unpaired electrons. Substituting $n=4$, $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90\text{BM}$. Thus, option (b) is correct.

Solution

IUPAC name of element : 119 : ununennium

Solution

Electronic configuration of Gadolinium Gd :— [Xe] 4f7 5d1 6s2 In case of 3 rd ionisation enthalpy electron will be removed from 5d and resultant configuration will be [Xe]4f7 that is stable electronic configuration as it will have high exchange energy, hence less energy will be required to remove 3rd electron. - 24 - NEET (UG)-2022 (Code-Q1)

Solution

• In an atom, all the five 3d orbitals are equal in energy in free state i.e., degenerate. • The shape of 22−xyd is different then shape of 2zd • The size of orbital depends on principal quantum number ‘n’ therefore all the five 3 d orbitals are different in size when compared to the respective 4d orbitals. • Shape of orbitals depends on azimuthal quantum number ‘l’ therefore shapes of 4 d orbitals are similar to the respective 3d orbitals.

Solution

21 xx 2 2 2 2 2 z 2 yy 2p * 2p *1 1 2 * 2 2 | | 2p 2p s s s s p ππ σ σ σ σ σ ππ Due to one unpaired electron in * 2pπ molecular orbital, 2O+ is a paramagnetic ion. SECTION-B

Solution

2 n nr Z∝ 22 33 22 22 r (Li ) (n ) Z(He ) r (He ) Z(Li ) (n ) + + ++ =× 2 2 3 2 r (Li ) (3) 2 105.8 3 (2) + =× 3105.8 2=× 2 3r (Li ) 158.7 pm+ =

Solution

The radius of the $n$-th orbit in a hydrogen atom is given by $r_n=n^2\cdot r_1$, where $r_1=5.3\times 10^{-11}\text{m}$. For the third orbit, $r_3=3^2\cdot 5.3\times 10^{-11}=4.77\times 10^{-10}\text{m}=4.77\text{A˚}$, so option (a) is correct.

Solution

The number of permissible values of the magnetic quantum number $m$ for a given value of the azimuthal quantum number $l$ is $n_m=2l+1$. Rearranging, $l=\frac{n_m-1}{2}$, so option (b) is correct.

Solution

Statement B is correct as the mass of the electron is indeed $9.10939\times 10^{-31}\text{kg}$. Statement C is correct because isotopes of an element have the same number of electrons and thus show the same chemical properties. Statement E is correct as Dalton’s atomic theory considered the atom to be an indivisible particle. Statements A and D are incorrect; atoms are composed of three fundamental particles (protons, neutrons, and electrons), and nucleons refer only to protons and neutrons. Therefore, option (a) is correct.

Solution

Statement I is true as atoms are electrically neutral because they contain equal number of positive and negative charges. Statement II is wrong as atom of most of the elements are stable and emit characteristic spectrum. But this statement is not true for every atom.

Solution

• Magnetic quantum number ml informs about orientation of orbital. ? • Spin quantum number ms informs about orientation of spin of electron. • Azimuthal quantum number (l) informs about shape of orbital ? • Principal quantum number (n) informs about size of orbital

Solution

Sol. $\Delta E=\frac{hc}{\lambda }=E_{\text{final}}-E_{\text{initial}}\left(E_n=\frac{-R_H}{n^2}\right)$

$\Delta E_{2\to 3}=\frac{hc}{{\lambda }_{2\to 3}}=E_3-E_2=\frac{-R_H}{3^2}-\left(\frac{-R_H}{2^2}\right)$

$=R_H\left(\frac{1}{4}-\frac{1}{9}\right)$

$=R_H\times \frac{5}{36}$

$\therefore {\lambda }_{2\to 3}=\frac{hc\cdot 36}{R_H\cdot 5}$

$\Delta E_{4\to 6}=E_6-E_4=\frac{-R_H}{36}+\frac{R_H}{16}=\frac{R_H\times 20}{36\times 16}$

$\frac{hc}{{\lambda }_{4\to 6}}=\frac{R_H\times 20}{36\times 16}$

${\lambda }_{4\to 6}=\frac{hc\times 36\times 16}{R_H\cdot 20}$

$\frac{{\lambda }_{2\to 3}}{{\lambda }_{4\to 6}}=\frac{\frac{hc\cdot 36}{R_H\cdot 5}}{\frac{hc\times 36\times 16}{R_H\cdot 20}}$

$=\frac{1}{4}$

Solution

Dalton’s theory could explain the laws of chemical combination. However, it could not explain the laws of gaseous volumes.

Solution

A. [NiCl₄]²⁻: Ni²⁺, 3d⁸, sp³ hybridisation, 2 unpaired electrons → paramagnetic.

B. Ni(CO)₄: Ni, 3d⁸ 4s², sp³ hybridisation, zero unpaired electron → diamagnetic.

C. [Ni(CN)₄]²⁻: Ni²⁺, 3d⁸, dsp² hybridisation, zero unpaired electron → diamagnetic.

D. [Ni(H₂O)₆]²⁺: Ni²⁺, 3d⁸, sp³d² hybridisation, two unpaired electrons → paramagnetic.

E. Ni(PPh₃)₄: Ni, 3d⁸ 4s², sp³ hybridisation, zero unpaired electron → diamagnetic.

Thus, A and D are paramagnetic.