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Current Question (ID: 9130)
Question:
$\text{The incorrect statement among the following is-}$
Options:
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1. $\text{Alkali metals are more electropositive than alkaline earth metals.}$
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2. $\text{Alkaline earth metals are denser and harder than alkali metals.}$
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3. $\text{An alkali metal's electronic configuration is ns}^1 \text{ and an alkaline earth metal's electronic configuration is ns}^2$
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4. $\text{The first IP of alkali metals is more than that of alkaline earth metals.}$
Solution:
$\text{HINT: Alkaline earth metals are denser than alkali metals.}$ $\text{Explanation:}$ $\text{1.) Alkali metals are more electropositive than alkaline earth metals. Alkali metals have only one electron in their valence shell. They can readily lose such an electron and attain the electronic configuration of the nearest noble gas.}$ $\text{2.) Because of their small size, alkaline earth metals are more closely packed in the crystal lattice. So they have comparatively stronger metallic bonds and so denser and harder than alkali metals.}$ $\text{The density values are as follows:}$ $\text{Li 0.53 ; Na=0.97 ; K= 0.86; Rb =1.53; Cs =1.90}$ $\text{Be = 1.84 ; Mg = 1.74 ; Ca = 1.55 ; Sr = 2.63}$ $\text{The strength of metallic bonds in alkaline earth metals is higher than alkali metals due to the presence of 2 electrons in its outermost shell as compared to alkali metals, which have only 1 electron in the valence shell. Therefore, alkaline earth metals are harder than alkali metals.}$ $\text{3.) Along the period from left to right, atomic number increases and, therefore, nuclear charge also increases.}$ $\text{So first ionization energies of alkaline earth metals are higher than those of the alkali metals of the same period because of higher nuclear charge.}$
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