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Question

The correct statement for alkali metal is:

The correct answer is
Reducing power increases from top to bottom

Alkali Metals: Understanding Reducing Power Trend

Alkali metals are the elements found in Group 1 of the periodic table. This group includes Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs), and Francium (Fr). A key characteristic of these metals is their high reactivity, which stems from having a single electron in their outermost energy shell (general electronic configuration: $ns^1$).

Defining Reducing Power

Reducing power is a measure of how easily a substance can donate electrons to another substance. A substance with high reducing power is easily oxidized itself. For alkali metals, this means they readily lose their single valence electron to form a positive ion (cation), typically $M^+$. The reaction is:

$M \rightarrow M^+ + e^-$

Analyzing Trends Down the Group

To understand the trend in reducing power, let's look at related properties:

  • Atomic Radius: As you move down the group from Lithium to Francium, the atomic radius increases significantly. This is because each subsequent element has an additional electron shell.
  • Ionization Energy ($IE$): The energy required to remove the outermost electron (the first ionization energy, $IE_1$) decreases down the group. This is a direct consequence of the increasing atomic size. The valence electron is further from the nucleus and is shielded by more inner electrons, making it easier to remove. The general trend is: $IE_1(\text{Li}) > IE_1(\text{Na}) > IE_1(\text{K}) > IE_1(\text{Rb}) > IE_1(\text{Cs})$.
  • Ease of Oxidation: Since the ionization energy decreases down the group, it becomes progressively easier for the alkali metal atoms to lose their valence electron.

Because it becomes easier to lose an electron as you go down the group, the reducing power of alkali metals increases from top to bottom.

Oxidising Power Consideration

Conversely, alkali metals have very little tendency to gain electrons. Therefore, they are very weak oxidizing agents. While their standard electrode potentials ($E^0$) show some variation, their primary chemical characteristic is their strong reducing nature.

Conclusion on the Correct Statement

Based on the decreasing ionization energy and increasing atomic size down the group, the ability of alkali metals to lose electrons (their reducing power) increases. Therefore, the correct statement is that the reducing power increases from top to bottom.

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Important Questions from p - Block

  1. Which of the following has the lowest boiling point?

  2. Which of the following reaction(s) do(es) NOT occur

    (i) [NPCl2]3 + 6 NaF \(\rm \xrightarrow[reflux]{MeCN}\) [NPF2]3 + 6 NaCl

    (ii) n PCl5 + n NH4Cl \(\rm \xrightarrow[reflux]{C_6H_5Cl}\) [NPCl2]n + 4 n HCl [n = 3, 4, 5]

    (iii) n PF 5  + n NH 4 F  \(\rm \xrightarrow[reflux]{C_6H_5Cl}\)  [NPF 2 ] n  + 4 n HF [n = 3, 4, 5]

  3. Choose the correct statement(s) from the following:

    (i) The trend in Lewis acidity among silicon halides is SiI4 < SiBr4 < SiCl4 < SiF4.

    (ii) Tin(II) chloride can act as a Lewis acid and not as a Lewis base.

    (iii) Aluminosilicates can display Brønsted acidity.

  4. Consider following statements

    A. PbCl2 has low solubility in water.

    B. Sulfides of As(III) and Sb(III) are soluble in ammonium sulfide.

    C. SnS is soluble in yellow ammonium sulfide.

    D. MnS is precipitated by passing H2S through acidic MnCl2.

    Correct statements are

  5. Which of the statements (A‐D) given below are correct for B2H6 molecule:

    A. Addition of Et2O•BF3 to NaBH4 in a polyether solvent produces B2H6.

    B. It has D2d symmetry.

    C. Reaction of B2H6 with NMe3 gives Me3N•BH3.

    D. It is diamagnetic.

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