Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N.
Choose the correct answer from the options given below:
Be > B (filled 2s); N > C (half-filled 2p). Order: Li < B < Be < C < N.
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Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N.
Choose the correct answer from the options given below:
Be > B (filled 2s); N > C (half-filled 2p). Order: Li < B < Be < C < N.
Among Group 16 elements, which one does NOT show $-2$ oxidation state?
Po is the most metallic Group 16 element and does not exhibit a stable $-2$ state.
In which of the following equilibria, $K_p$ and $K_c$ are NOT equal?
$K_p = K_c(RT)^{\Delta n}$. Only (B) has $\Delta n = 1 \ne 0$.
The reagents with which glucose does not react to give the corresponding tests/products are:
A. Tollen's reagent B. Schiff's reagent C. HCN D. NH₂OH E. NaHSO₃
Choose the correct options from the given below:
Glucose exists largely in cyclic hemiacetal form, so its aldehyde group is masked — it does not give a positive Schiff's test or react with NaHSO₃.
Identify the correct reagents that would bring about the following transformation.
Ph–CH₂–CH=CH₂ → Ph–CH₂–CH₂–CHO
Anti-Markovnikov OH placement needs hydroboration-oxidation, then PCC oxidises the primary alcohol selectively to the aldehyde (CrO₃ would over-oxidise to the carboxylic acid).
The $E^\circ$ value for the $\text{Mn}^{3+}/\text{Mn}^{2+}$ couple is more positive than that of $\text{Cr}^{3+}/\text{Cr}^{2+}$ or $\text{Fe}^{3+}/\text{Fe}^{2+}$ due to change of:
Mn³⁺ ($d^4$) → Mn²⁺ ($d^5$, half-filled, stable) — large favourable change, hence more positive $E^\circ$.
Given below are two statements:
Statement I: Both $[\text{Co(NH}_3)_6]^{3+}$ and $[\text{CoF}_6]^{3-}$ complexes are octahedral but differ in their magnetic behaviour.
Statement II: $[\text{Co(NH}_3)_6]^{3+}$ is diamagnetic whereas $[\text{CoF}_6]^{3-}$ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
NH₃ is strong field → low-spin $d^6$ (no unpaired e⁻, diamagnetic). F⁻ is weak field → high-spin $d^6$ (4 unpaired e⁻, paramagnetic).
Fehling's solution 'A' is
Fehling A is aqueous CuSO₄; Fehling B is alkaline sodium-potassium tartrate.
In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C. $2\,\text{NaHCO}_3{}_{(s)} \to \text{Na}_2\text{CO}_3{}_{(s)} + \text{CO}_2{}_{(g)} + \text{H}_2\text{O}_\text{(g)}$
D. $\text{Cl}_2{}_{(g)} \to 2\,\text{Cl}_{(g)}$
Choose the correct answer from the options given below:
Evaporation (A), decomposition producing gases (C) and atomisation (D) all increase entropy; cooling a crystal (B) decreases entropy.
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.
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