NEET Chemistry Questions: Classification of elements and Periodicity in properties

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The most electronegative element possess the electronic configuration




The most electronegative element is fluorine, which has the electronic configuration $1s^2 2s^2 2p^5$. This matches the general form $ns^2 np^5$, where n=2 for fluorine. Therefore, the correct option is $ns^2 np^5$.
The ionisation enthalpy of Cs is $ 375.6KJmol^{–1} $ what is the energy required to convert [at mass of Cs = 133] 2.66mg of gaseous Cs completely to $ Cs^+ $




First, we need to convert the mass of Cs to moles: \[ \text{Moles of Cs} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.66 \text{ mg}}{133 \text{ g/mol}} \] Since 1 mg = 0.001 g, \[ \text{Moles of Cs} = \frac{2.66 \times 10^{-3} \text{ g}}{133 \text{ g/mol}} \approx 2 \times 10^{-5} \text{mol} \] The ionization enthalpy of Cs is given as 375.6 kJ/mol. Therefore, the energy required to convert 2.66 mg of Cs to \( Cs^+ \) is: \[ \text{Energy} = \text{moles} \times \text{ionization enthalpy} = 2 \times 10^{-5} \text{ mol} \times 375.6 \text{ kJ/mol} \] \[ \text{Energy} = 7.512 \text{ J} \] Thus, the correct option is 7.512 J.
The atomic number of elements M, N, & P are x, x–1, x–3. If P is a halogen atom then the type of bond between N & P is




If P is a halogen (from Group 17), it has 7 valence electrons. Since N has an atomic number of x-1, it would most likely be an alkali metal with 1 valence electron. The bond between an alkali metal and a halogen is typically ionic due to the transfer of the electron from the metal to the halogen. Therefore, the bond between N and P is ionic.
An element X belongs to Gp16 & 5th period. Its atomic number is




An element in Group 16 and the 5th period of the periodic table is Tellurium (Te), which has an atomic number of 52. Thus, the correct option is 52.
The position of an element with atomic number 114 is




Element with atomic number 114 is known as Flerovium (Fl). It belongs to the 7th period and group 14 of the periodic table.
The size of Mo is very similar to W due to




The size of Molybdenum (Mo) is very similar to Tungsten (W) due to the poor shielding effect by 4f electrons. The poor shielding results in a higher effective nuclear charge, causing the atomic radii to be similar.

Choose the correct order ionization energy





No explanation available.
The order of ionization energy of K, Ca, & Ba are




The order of ionization energy for the elements K (Potassium), Ca (Calcium), and Ba (Barium) can be explained based on their positions in the periodic table. Ionization energy typically increases across a period and decreases down a group. Calcium (Ca) is in the same group as Barium (Ba) but is above it, so Ca has a higher ionization energy than Ba. Potassium (K) is in a different group, and its ionization energy is less than that of both Ca and Ba. Thus, the correct order is $Ca > Ba > K$.
The element with zero electron gain enthalpy is




Argon (Ar) is a noble gas with a completely filled valence shell, making it very stable and not inclined to gain any electrons. Thus, it has zero electron gain enthalpy. Lithium (Li), Calcium (Ca), and Fluorine (F) do not have zero electron gain enthalpy as they can gain electrons to achieve a stable configuration.
Pick the iso electronic species from the following I. $NH_3$ II. $NH^– _ 2 $ III. $CH^+ _3 $ IV. $H_3 O^+ $




No explanation available.
The element with atomic number 44 belongs




The element with atomic number 44 is Ruthenium (Ru), which belongs to the d-block of the periodic table. Elements in the d-block are also known as transition metals.
In the third period there are only eight elements because




No explanation available.
Choose the correct electronic configuration which has the highest difference between first & second ionisation enthalpies.




The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$ corresponds to a noble gas core configuration with one electron in the 4s orbital. The first ionization energy removes this 4s electron, which is relatively easy. However, the second ionization energy requires removing an electron from the stable, noble gas core, which is significantly more difficult. This results in a high difference between the first and second ionization enthalpies.
The set of quantum numbers for unpaired electron of an element with atomic number 84 are




No explanation available.
The elements with highest ionization enthalpy in a period are




Noble gases have the highest ionization enthalpy in a period because they have completely filled valence shells, making them very stable and requiring a significant amount of energy to remove an electron. Therefore, option o3 is correct.