मराठी

Draw the Electron-dot Structure Of Hcl Compound and State the Type of Bonding.

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प्रश्न

Draw the electron-dot structure of  HCl  compound and state the type of bonding.

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उत्तर

The electron-dot structure of HCl is:

The hydrogen atom shares its one electron with a chlorine atom to form a covalent bond.

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पाठ 3: Metals and Non-metals - Exercise 2 [पृष्ठ १६९]

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लखमीर सिंह Chemistry [English] Class 10
पाठ 3 Metals and Non-metals
Exercise 2 | Q 38.4 | पृष्ठ १६९

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

List three characteristic properties of covalent compounds.


State the reason why carbon can neither form C4+ cations nor C4− anions but forms covalent compounds.


State whether the following statement is true or false:

Diamond and graphite are the covalent compounds of carbon element (C)


Which inert gas does the Cl atom in HCl resemble in electron arrangement?


Draw the electron-dot structure of H2O compound and state the type of bonding.


State any two uses of diamond.


Out of KCl, HCl, CCl4 and NaCl, the compounds which are not ionic are:
(a) KCl and HCl
(b) HCl and CCl4
(c) CCl4 and NaCl
(d) KCl and CCl4


The number of carbon atoms joined in a spherical molecule of buckminsterfullerene is:
(a) fifty
(b) sixty
(c) seventy
(d) ninety


Give the formulae of the chlorides of the elements X and Y having atomic numbers of 3 and 6 respectively. Will the properties of the two chlorides be similar or different? Explain your answer.


Explain, giving reason, why carbon neither forms C4+ cations nor C4− anions, but forms covalent compounds which are bad conductors of electricity and have low melting point and low boiling point. 

Explain the Structural isomerism term with example.


Explain the following:
Non-polar covalent compounds are insoluble in water.


Give reason as to why hydrogen chloride can be termed as a polar covalent compound.


Explain the bonding in methane molecule using electron dot structure.


Element A has 2 electrons in its M shell. Element B has atomic number 7.

If B is a diatomic gas, write the equation for the direct combination of A and B to form a compound.


Give an example for each of the following statement.

Formation of coordinate covalent bond between NH3 ➝ BF3 molecules


Identify the incorrect statement and correct them.

  1. Like covalent compounds, coordinate compounds also contain charged particles (ions). So they are good conductors of electricity. 
  2. Ionic bond is a weak bond when compared to Hydrogen bond.
  3. Ionic or electrovalent bonds are formed by mutual sharing of electrons between atoms. 
  4. Loss of electrons is called Oxidation and gain of electron is called Reduction.
  5. The electrons which are not involved in bonding are called valence electrons.

Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, e.g., hydrogen. After the formation of four bonds, carbon attains the electronic configuration of ______.


Which of the following are correct structural isomers of butane?

  1. \[\begin{array}{cc}
    \ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
    |\phantom{....}|\phantom{....}|\phantom{....}|\\
    \ce{H - C - C - C - C - H}\\
    |\phantom{....}|\phantom{....}|\phantom{....}|\\
    \ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
    \end{array}\]

  2. \[\begin{array}{cc}
    \ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
    |\phantom{....}|\phantom{....}|\\
    \ce{H - C - C - C - H}\\
    |\phantom{.....}|\phantom{.....}|\\
    \ce{H}\ce{H-C-H}\ce{H}\\
    |\\
    \ce{H}\\
    \end{array}\]
  3. \[\begin{array}{cc}
    \ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
    |\phantom{....}|\phantom{....}|\\
    \ce{H - C - C - C - H}\\
    |\phantom{.....}\backslash\phantom{..}|\\
    \phantom{....}\ce{H}\phantom{......}\ce{C - H}\phantom{}\\
    \phantom{.......}|\\
    \phantom{.......}\ce{H}\\
    \end{array}\]
  4. \[\begin{array}{cc}
    \ce{H}\phantom{...}\ce{H}\\
    |\phantom{....}|\\
    \ce{H - C - C - H}\\
    |\phantom{....}|\\
    \ce{H - C - C - H}\\
    |\phantom{....}|\\
    \ce{H}\phantom{...}\ce{H}\\
    \end{array}\]

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