Among the given elements: O, Cl, F, N, P, Sn, Tl, Na and Ti, the number of elements showing only one non-zero oxidation state is:
(A) 4
(B) 2
(C) 3
(D) 5
Answer
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Hint: A more practical method of using oxidation numbers to keep track of electron shifts in a chemical reaction involving the formation of many covalent compounds. Consider there is a complete transfer of electrons from a less electronegative atom to a more electronegative atom in this method. In a chemical reaction assume that electron transfer leads to the description of redox reactions by using oxidation numbers.
Complete answer:
Oxidation number represents the oxidation state of an element in a compound established based on a set of rules formulated on an electron pair in a covalent bond that belongs entirely to a more electronegative element.
The charge on an atom appears to have where all other atoms are removed from its ions are called oxidation numbers, which may have +Ve or –Ve sign with a whole number or fractional or zero. Depending on the nature of the compound present, an element has different values of oxidation number. O, Cl, F, N, P, Sn, Tl, Na, and Ti
The oxidation states of Oxygen (O) = -1, +1, -2
The oxidation states of chlorine (Cl) = -1, +1, +2, +3, +4, +5, +6, +7
The oxidation state of fluorine (F) = -1
The oxidation states of nitrogen (N) = -3, -2, -1, +1, +2, +3, +4, +5
The oxidation states of Phosphorous (P) = -3, -2, -1, +1, +2, +3, +4, +5
The oxidation states of tin (Sn) = +2, +3, +4
The oxidation state of Thallium (Tl) = +1, +3
The oxidation state of sodium (Na) = +1
The oxidation state of titanium (Ti) = +2, +3, +4
Hence, the number of elements O, Cl, F, N, P, Sn, Tl, Na, and Ti showing only one non-zero
oxidation states of two elements - Na (+1), and F (-1).
The correct answer is option B.
Note:
There are some rules for determining the oxidation number of ions, or molecules, or compounds. In the elementary state of all elements has oxidation number Zero. Fluorine is the most electronegative element of all elements, which shows only -1 oxidation number in all of its compounds. The lowest and highest number of oxidation numbers in transition elements are equal to ns electrons and ‘ns’, (n-1) d unpaired electrons respectively.
Complete answer:
Oxidation number represents the oxidation state of an element in a compound established based on a set of rules formulated on an electron pair in a covalent bond that belongs entirely to a more electronegative element.
The charge on an atom appears to have where all other atoms are removed from its ions are called oxidation numbers, which may have +Ve or –Ve sign with a whole number or fractional or zero. Depending on the nature of the compound present, an element has different values of oxidation number. O, Cl, F, N, P, Sn, Tl, Na, and Ti
The oxidation states of Oxygen (O) = -1, +1, -2
The oxidation states of chlorine (Cl) = -1, +1, +2, +3, +4, +5, +6, +7
The oxidation state of fluorine (F) = -1
The oxidation states of nitrogen (N) = -3, -2, -1, +1, +2, +3, +4, +5
The oxidation states of Phosphorous (P) = -3, -2, -1, +1, +2, +3, +4, +5
The oxidation states of tin (Sn) = +2, +3, +4
The oxidation state of Thallium (Tl) = +1, +3
The oxidation state of sodium (Na) = +1
The oxidation state of titanium (Ti) = +2, +3, +4
Hence, the number of elements O, Cl, F, N, P, Sn, Tl, Na, and Ti showing only one non-zero
oxidation states of two elements - Na (+1), and F (-1).
The correct answer is option B.
Note:
There are some rules for determining the oxidation number of ions, or molecules, or compounds. In the elementary state of all elements has oxidation number Zero. Fluorine is the most electronegative element of all elements, which shows only -1 oxidation number in all of its compounds. The lowest and highest number of oxidation numbers in transition elements are equal to ns electrons and ‘ns’, (n-1) d unpaired electrons respectively.
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