Classification of elements and Periodicity in properties MCQs for NEET — Chemistry Questions with Answers

Maxmimum density of water is at $ 40 ^\circ C $ also $ { C \over 5} = { F -32 \over 9 } $

Carbon cannot form $ C^{+4} $ or $ C^{-4} $ ions mainly due to the high ionization enthalpy and high electron gain enthalpy required. Ionization enthalpy refers to the energy required to remove electrons, and for carbon to lose four electrons, it would require a very high amount of energy. Similarly, electron gain enthalpy is the energy change when an electron is added to an atom. Adding four electrons to carbon would also require a significant amount of energy, making it highly unfavorable. Therefore, both high ionization enthalpy and high electron gain enthalpy prevent the formation of $ C^{+4} $ or $ C^{-4} $ ions.

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 $ns^2 np^5$.

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.

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 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.

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 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.

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 $Ca > Ba > K$.