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NEET UG Unit 1C1: Periodic Trends in Atomic Properties — practice questions

Explore 4 free Unit 1C1: Periodic Trends in Atomic Properties questions in NEET UG Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1C1: Periodic Trends in Atomic Properties questions and exercises with answers

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(i) $\mathrm{A} + \mathrm{Na}_{2}\mathrm{CO}_{3} \longrightarrow \mathrm{B} + \mathrm{C}$ (ii) A $\xrightarrow{\mathrm{CO}_2}$ (Milky) C The chemical formula of A and B are respectivelyRead the answer and worked solution →Which process of purification is represented by the following scheme $\mathrm{Ti}_{\text{impure}} + 2\mathrm{I}_2 \xrightarrow{250^\circ \mathrm{C}} \mathrm{TiI}_4 \xrightarrow{1400^\circ \mathrm{C}} \mathrm{Ti}_{\text{pure}} + \mathrm{I}_2$Read the answer and worked solution →Solid $PCl_5$ exist as:Read the answer and worked solution →NEET · Chemistry · Alkali Metals — Property MatchRead the answer and worked solution →

Worked practice questions

(i) $\mathrm{A} + \mathrm{Na}_{2}\mathrm{CO}_{3} \longrightarrow \mathrm{B} + \mathrm{C}$ (ii) A $\xrightarrow{\mathrm{CO}_2}$ (Milky) C The chemical formula of A and B are respectively

  1. NaOH and $\mathrm{Ca(OH)_2}$
  2. $\mathrm{Ca(OH)_2}$ and NaOH
  3. NaOH and CaO
  4. CaO and $\mathrm{Ca(OH)_2}$
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Correct answer: B. $\mathrm{Ca(OH)_2}$ and NaOH

From statement $\mathrm{A} \xrightarrow{\mathrm{CO}_2} \mathrm{C}$ with $\mathrm{C}$ described as milky, $\mathrm{A}$ must be lime water and $\mathrm{C}$ must be the white precipitate formed when carbon dioxide is passed through it. $$\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O}$$ So $\mathrm{A = Ca(OH)_2}$ and $\mathrm{C = CaCO_3}$. Now use equation $(i)$. $$\mathrm{Ca(OH)_2 + Na_2CO_3 \rightarrow 2NaOH + CaCO_3}$$ Hence $\mathrm{B = NaOH}$. So the pair is $\mathrm{A = Ca(OH)_2}$ and $\mathrm{B = NaOH}$. Rechecking the options, this matches option $\mathrm{(B)}$.

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Which process of purification is represented by the following scheme $\mathrm{Ti}_{\text{impure}} + 2\mathrm{I}_2 \xrightarrow{250^\circ \mathrm{C}} \mathrm{TiI}_4 \xrightarrow{1400^\circ \mathrm{C}} \mathrm{Ti}_{\text{pure}} + \mathrm{I}_2$

  1. Van-Arkel process
  2. Cupellation
  3. Electrolytic refining
  4. Zone refining
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Correct answer: A. Van-Arkel process

This purification involves formation of a volatile iodide of titanium at lower temperature and its decomposition at high temperature to deposit pure metal. That is the iodide process, also called the Van-Arkel process. The given reaction is: $$\mathrm{Ti}_{\text{impure}} + 2\mathrm{I}_2 \xrightarrow{250^\circ \mathrm{C}} \mathrm{TiI}_4$$ $$\mathrm{TiI}_4 \xrightarrow{1400^\circ \mathrm{C}} \mathrm{Ti}_{\text{pure}} + \mathrm{I}_2$$ Among the options, this matches $\text{Van-Arkel process}$.

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Solid $PCl_5$ exist as:

  1. $[PCl_4]^+$ and $[PCl_6]^-$
  2. $PCl_3$ and $Cl_2$
  3. $[PCl_4]^-$ and $[PCl_6]^+$
  4. $[PCl_3]^+$ and $[PCl_3]^-$
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Correct answer: A. $[PCl_4]^+$ and $[PCl_6]^-$

In the solid state, $PCl_5$ is ionic and exists as $[PCl_4]^+$ and $[PCl_6]^-$ ions. This arrangement stabilizes the compound in the crystal lattice. **Hence option A is correct.**

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