The compound which does not exist.
- BeCl2
NaO2
- PbEt4
- (NH4)2 BeF4
The Correct Option is B
Approach Solution - 1
BeCl2, PbEt4, and (NH4)2BeF4 are all valid chemical compounds. BeCl2 is beryllium chloride, PbEt4 is tetraethyllead, and (NH4)2BeF4 is ammonium beryllium fluoride.
Answer. B
Approach Solution -2
The correct answer is NaO2
Lithium forms oxide, sodium forms peroxide and the K, Rb, Cs forms superoxide.
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Reasons R: The deep blue solution to the formation of amide.
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B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
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Concepts Used:
Group 1 Elements
Group one of alkali metals is s-block elements with just one electron in their s-orbital. They are are alkali metals. They are named so because of the alkaline nature of the hydroxides and oxides.
Alkali metals are characterized by one s-electron in the valence shell of their atoms.
Alkali metals have a corresponding [Noble gas] ns1 electronic configuration. They occupy the first column of the periodic table. Alkali elements are:
- Lithium(Li)
- Sodium(Na)
- Potassium (K)
- Rubidium (Ru)
- Cesium (Cs)
- Francium (Fr)
They have occupied successive periods from first to seven. Francium is a radioactive element with very low half-life.
Electronic Configuration:
- Alkali metals have one electron in their valence shell.
- The electronic configuration is given by ns1. For example, the electronic configuration of lithium is given by 1ns1 2ns1.
- They tend to lose the outer shell electron to form cations with charge +1 (monovalent ions).
This makes them the most electropositive elements and due to the same reason, they are not found in the pure state.
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