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Question

__________ nitrides are termed by transition + inner transition metals.

The correct answer is
Interstitial

To answer the given question, we will examine the nature of nitrides formed by transition and inner transition metals.

Nitrides are compounds that consist of nitrogen combined with a more electropositive element. Generally, nitrides can be classified into three types:

  1. Salt-like Nitrides: These are ionic compounds formed with alkali and alkaline earth metals. They are characterized by their salt-like structure.
  2. Covalent Nitrides: These nitrides are formed with elements that have similar electronegativities to nitrogen, such as boron or silicon. They have a covalent bond structure.
  3. Interstitial Nitrides: These are typically formed by transition and inner transition metals. The nitrides have nitrogen atoms occupying the interstitial sites in the metal lattices. This type of nitride exhibits metallic properties and is known for being stable and hard.

Now, considering the options and definitions, the classification that best matches the type of nitrides formed by transition and inner transition metals is:

  • Salt-like: Incorrect, because salt-like nitrides are not typically formed by transition metals.
  • Covalent: Incorrect, as these involve more covalent character, rather than the typical formation by transition metals.
  • Diamond-like: Incorrect, this is not a standard classification of nitrides.
  • Interstitial: Correct, because transition and inner transition metals often form interstitial nitrides where nitrogen occupies spaces in the metal lattice.

Thus, the correct answer is Interstitial.

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Important Questions from Main Group Elements and Their Compounds

  1. A non-metal ‘X’ occurs in nature in free state. It combines with hydrogen to form a gas, which is colourless and highly toxic. Apart from this, ‘X’ is oxidized to ‘$\text{XO}_2$’ which is a colourless, pungent and irritating gas. ‘X’ represents :
  2. For the following reaction sequence,


    X and Y, respectively, are

  3. The reaction of $PCl_3$ with $PhLi$ in 1:3 molar ratio yields X as one of the products, which on further treatment with $CH_3I$ gives Y. The reaction of Y with $n-BuLi$ gives product Z. The products X, Y and Z, respectively, are

  4. Conversion of boron trifluoride to tetrafluoroborate accompanies
  5. The correct statements regarding B among the following

    (i) Nuclear spin of 11B is greater than that of 10B

    (ii) The polarities of B - H bond and C-H bonds are opposite

    (iii) Cross - section of neutron absorption for 10B is much more than that of 11B

    (iv) B reacts with boiling aq. NaOH solution to form NaB(OH)4

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