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The s-Block Elements Test 0

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The s-Block Elements Test 0
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which one of the following species does not exist under normal conditions?
    Solution
    $$Be_{2}$$ does not exist under normal condition. Beryllium has four electrons , so electronic configuration of $$Be_{2}$$ having 8 electrons will be $$ \sigma 1s^{2} \sigma^{*}1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2}$$ and the bond order is 
    $$ \cfrac{N_{a} - N_{b}}{2} = 0$$
    Since $$Be_{2}$$ is having bond order 0 , no bond is formed between the two atoms. Hence $$Be_{2}$$ is unstable 
  • Question 2
    1 / -0
    Mg is present in 
    Solution
    Structure of Chlorophyll  - It contains Mg  

    NOTE-  Haemoglobin contains Fe

  • Question 3
    1 / -0
    Sulphate of an alkaline earth metal which crystalizes without water of hydration is?

    Solution
    Strontium sulphate crystalizes without water of hydration.
  • Question 4
    1 / -0
    Sodium and potassium react with water much more vigorously than lithium because:
    Solution
    When sodium and potassium react with water, the heat evolved causes them to melt, giving a larger area of contact with water, lithium on the other hand, does not melt under these condition and thus reacts more slowly.

    Li      Na     K
    180   98    64   Melting point $$(^oC)$$
  • Question 5
    1 / -0
    A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water, it forms a solution which turns red litmus blue. Identify X :
    Solution
    Substance X is calcium oxide $$(CaO)$$, also known as quicklime. It is an important ingredient in the processing of cement. Calcium oxide reacts with water to form calcium hydroxide $$Ca(OH)_{2}$$ which is a base and therefore, turns red litmus blue. The equation can be written as:
    $$2CaO(s)  +  H_{2}O(l)  \rightarrow   Ca(OH)_{2}(aq)$$
  • Question 6
    1 / -0
    The colour of transition metal compound is due to :
    Solution

    Usually the color of transition metal compound is due to d-d transition. When the d-level is not completely filled , it is possible to promote electron from a lower energy d-orbital to higher energy d-orbital by absorption of a photon of electromagnetic radiation having an appropriate energy.

    Hence on the otherhand we can say that the color of transition metal is due to incomplete (n-1) d- subshell.

  • Question 7
    1 / -0
    Alkaline earth metal nitrates on heating decompose to give:
    Solution
    Alkaline earth metal nitrates on heating decompose to give metal oxide, nitrogen dioxide and oxygen gas. These are white solid and brown nitrogen dioxide and oxygen gas is given off when heated.
    $$2M(NO_3)_2 \rightarrow 2MO + 4NO_2 + O_2$$
  • Question 8
    1 / -0
    Consider the following statements and arrange in the order of true/false as given in the codes.
    $$S_1$$: Sodium amalgam reacts with hot water and produces sodium hydroxide and hydrogen gas.
    $$S_2$$: Sodium is more reactive than potassium.
    $$S_3$$: Alkaline earth metals dissolve in liquid ammonia to give deep blue-black solutions which are conducting.
    $$S_4$$: Alkaline earth metal salts have more number of water molecules as compared to those of alkali metal salts.
    Solution
    On moving down the group, the group atomic size increases and the effective nuclear charge decreases.
    Because of this factor, the outermost electron in potassium can be lost easily as compared to sodium. Hence, potassium is more reactive than sodium. S1,  S3, S4 are all true 
  • Question 9
    1 / -0
    Property of all the alkaline earth metals that increases with their atomic number is:
    Solution
    Down the group, the thermal stability of group 2 sulphates increases with their  atomic number. This is due to the fact that  higher the atomic number, larger the ionic radius and lesser the charge density. So it causes less distortion to nearby negative charge, so the bond formed between the cation and anion will be strong and stable.
    Rest all the given properties decreases while going down the group.
    Option B is correct.
  • Question 10
    1 / -0
    The energy of an electron in a particular orbit of single electron species of beryllium is the same as the energy of an electron in the ground state of a hydrogen atom. Identify the orbit of beryllium.
    Solution
    Energy of an electron in $$Be^{+3}$$ ion
    $$\displaystyle = \frac{-13.6}{n^2} \times 4^2 ev$$
    Energy of electron in ground state hydrogen atom $$=-13.6 ev$$
    $$\therefore \displaystyle \frac{-13.6}{n^2} \times 4^2 =-13.6$$
    $$\Rightarrow n^2 = 4^2$$
    $$\therefore \displaystyle \frac{-13.6}{4^2} \times 4^2 =-13.6$$
    we get -13.6 on both sides 
    $$\Rightarrow n = 4$$
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