Self Studies

The p-Block Elements Test - 17

Result Self Studies

The p-Block Elements Test - 17
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    One mole of magnesium nitride on the reaction with an excess of water gives : 
    Solution
    Hint: Nitride on reaction with water gives ammonia
    Step 1: Formula of magnesium nitride
    The formula for magnesium nitride will be $$Mg_3N_2$$

    Step 2: Recation of magnesium nitride with excess of water
    $$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$$

    Hence one mole of magnesium nitride on reaction with excess water gives two moles of ammonia.

    Final Step: So the correct answer is option $$C$$.
  • Question 2
    1 / -0
    The reaction of white phosphorus with aqueous $$NaOH$$ gives phosphine along with another phosphorus containing compound. The reaction type, the oxidation states of phosphorus in phosphine and the other product are respectively:
    Solution
    $$\overset{0}{P_4} + 3NaOH + 3H_2O \rightarrow \overset{-3}{P}H_3 + 3NaH_2\overset{+1}{P}O_2$$      

    The salt $$NaH_2PO_2$$ undergoes the following changes on heating.

    $$4NaH_2\overset{+1}{P}O_2 \rightarrow Na_4\overset{+5}{P_2}O_7 + 2\overset{-3}{P}H_3 + H_2O$$

    The reaction is a disproportionation reaction and the oxidation state of phosphorus are -3 (in phosphine) and +5 in the second product.

    Option C is correct.
  • Question 3
    1 / -0
    The reaction of $$\mathrm{P}_{4}$$ with X leads selectively to $$\mathrm{P}_{4}\mathrm{O}_{6}$$. The X is:
    Solution
    $$P_4+3O_{2}\overset{N_{2}}{\rightarrow}P_{4}O_{6}$$

    White phosphorus on reaction with a limited supply of oxygen gives lower oxide, $$P_4O_6$$. 

    Therefore, air $$(O_2 + N_2)$$ is a good source for the controlled supply of oxygen and the best choice for controlled oxidation of white phosphorus into lower oxide.

    Hence, option B is correct.
  • Question 4
    1 / -0
    Bleaching powder and bleach solution are produced on a large scale and is used in several house-hold products. The effectiveness of bleach solution is often measured by iodometry. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:
    Solution
    Bleaching powder is $$Ca(OCl)Cl$$.

    So, bleaching powder contains a salt of $$HOCl$$.

    $$Cl_2O$$ is an anhydride of $$HOCl$$.

    $$Cl_2O+H_2O\Rightarrow 2HOCl$$.

    Hence, option A is correct.
  • Question 5
    1 / -0
    Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of :
    Solution
    The decomposition of $$HNO_3$$ is represented by the equation:

    $$4HNO_{3}\rightarrow 2H_{2}O+4NO_{2}+O_{2}$$

    Due to the formation of nitrogen dioxide, it turns yellow-brown.

    Hence, option B is correct.
  • Question 6
    1 / -0
    The correct structure of tribromooctaoxide is: 
    Solution
    The correct structure is A
    $$O=\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}-\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}-\overset{O}{\overset{||}{\underset{O}{\underset{||}{Br}}}}=O$$
    Tribromooctaoxide
  • Question 7
    1 / -0
    Which one of the following order is correct for the bond dissociation enthalpy of halogen molecules?
    Solution
    Bond dissociation energy of halogen family decreases down the group as the size of atom increases.

    The decreasing order for the bond dissociation enthalpy of halogen molecules is  $$\displaystyle Cl_{2} \text {  57 kcal/mol} > Br_{2} \text {  45.5 kcal/mol} > F_{2} \text {  38 kcal/mol} > \text {  35.6 kcal/mol} I_{2}$$.

    A halogen molecule having larger atoms should have low dissociation energy and vice versa. Fluorine is an exception because of interelectronic repulsion is present in small atom fluorine.

    Hence, the correct option is $$B$$.
  • Question 8
    1 / -0
    For electron affinity of halogens which of the following is correct?
    Solution
    Electron affinity decreases in a group from top to bottom due to increase in size, so the expected order of electron affinity for  halogens is $$F > Cl > Br > I$$
    However, it is not so. The electron affinity of elements of $$III$$ period is higher than that of $$II$$ period because elements of $$II$$ period have a small size and greater electron density so the incoming electrons suffer a repulsive force. Thus the order of electron affinity is $$Cl > F > Br > I$$.
  • Question 9
    1 / -0
    Assertion: F is more electronegative than $$Cl$$.
    Reason: F has high electron affinity than $$Cl$$.
    Solution
    Here assertion is true but reason is false. F is more electronegative than chlorine, in fact it is most electronegative element of the periodic table but its electron affinity, the tendency to accommodate electrons, is less than the $$Cl$$ due to non availability of $$d$$-electrons.
  • Question 10
    1 / -0
    Assertion: The fluorine has lower reactivity.
    Reason : $$F - F$$ bond has low bond dissociation energy.
    Solution

    Fluorine have the highest electronegativity of all the elements because it has the largest nuclear charge and the least atomic shell, so it has the largest nuclear pull force. In other words, fluoride has the strongest ability to pull or catch an electron to form negative ion (that is, electronegativity).

    Bond dissociation energy of $$F_2$$  is fairly low due to the lone pair - lone pair repulsions of the two flourine atoms, which are placed so close to each other due to small size of the atom.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now