Practice free Atomic Structure (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Which of the following orbitals corresponds to n=4, l=0?
According to the table in the NCERT, for l=0, the orbital is denoted by 's'. So, n=4, l=0 corresponds to a 4s orbital.
What does the azimuthal quantum number (l) primarily define?
The NCERT text states, 'azimuthal quantum number. ‘l’ is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital.'
The pairing of electrons in d-orbitals starts with the entry of which electron?
The NCERT text says, 'Since there are three p, five d and seven f orbitals, therefore, the pairing of electrons will start in the p, d and f orbitals with the entry of 4th, 6th and 8th electron, respectively.' For d-orbitals, pairing starts with the 6th electron, as there are 5 d-orbitals, and each gets one electron before pairing begins.
The relation between $n_m$ ($n_m$ = the number of permissible values of magnetic quantum number $m$) for a given value of azimuthal quantum number ($l$), is:
$n_m = 2l + 1\Rightarrow l = (n_m - 1)/2$.
Select the correct statements:
A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is $9.10939 \times 10^{-31}$ kg. C. All the isotopes of a given element show same chemical properties. D. Protons and electrons are collectively known as nucleons. E. Dalton's atomic theory, regarded the atom as an ultimate particle of matter.
A wrong (3 particles: p, n, e). D wrong (nucleons = p + n, not p + e). B, C, E correct.
Match List I with List II.
| List I (Quantum Number) | List II (Information provided) |
|---|---|
| A. $m_l$ | I. shape of orbital |
| B. $m_s$ | II. size of orbital |
| C. $l$ | III. orientation of orbital |
| D. $n$ | IV. orientation of spin of electron |
Choose the correct answer from the options given below:
$m_l$: orientation; $m_s$: spin orientation; $l$: shape; $n$: size.
The energy of an electron in the ground state ($n = 1$) for $\text{He}^+$ ion is $-x$ J, then that for an electron in $n = 2$ state for $\text{Be}^{3+}$ ion in J is:
$E \propto Z^2/n^2$. He⁺: $4/1 = 4$. Be³⁺ at $n=2$: $16/4 = 4$. Ratio is 1, so energy is $-x$ J.
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \to n = 3$ and $n = 4 \to n = 6$ transitions, respectively, is:
$\dfrac{1}{\lambda}\propto\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$. For $2\to3$: $\tfrac{5}{36}$; for $4\to6$: $\tfrac{5}{144}$. $\dfrac{\lambda_{2\to3}}{\lambda_{4\to6}} = \dfrac{5/144}{5/36} = \dfrac{1}{4}$.
Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are [Given $R_H = 2.18\times10^{-18}$ J, $a_0 = 52.9$ pm]:
$E_n = -R_H\dfrac{Z^2}{n^2}$, $r_n = a_0\dfrac{n^2}{Z}$. Li²⁺ ($Z=3$): $E = -2.18\times9 = -19.62\times10^{-18}$ J, $r = 52.9/3 = 17.6$ pm. He⁺ ($Z=2$): $E = -8.72\times10^{-18}$ J, $r = 26.4$ pm.
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