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The maximum valency of an element having atomic number seven is: (a) (b) 3 (c) 5 (d) 7

Short Answer

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(c) 5

Step by step solution

01

Identify the Element

The element with atomic number 7 is Nitrogen (N). This can be found on the periodic table.
02

Determine Possible Oxidation States

Nitrogen can have multiple oxidation states, with the most common being -3, +3, and +5. The valency is determined by the maximum positive oxidation state.
03

Find the Maximum Valency

The maximum valency of an element refers to the highest number of chemical bonds it can form, which corresponds to its highest oxidation state. For nitrogen, the highest oxidation state is +5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nitrogen Basics
Nitrogen, represented by the symbol \( \text{N} \), is the seventh element in the Periodic Table, with an atomic number of 7. It's one of the most common elements in the universe and makes up about 78% of the Earth's atmosphere. Nitrogen is a non-metal and belongs to the group of elements called "pnictogens." This group is characterized by elements that often form multiple bonds and many of these elements interact actively with oxygen.
  • Atomic Number: 7
  • Symbol: \( \text{N} \)
  • Group: Pnictogens (Group 15)
  • Non-metallic properties
Nitrogen’s ability to form multiple bonds makes it highly versatile in chemical reactions. It is commonly found in both organic and inorganic compounds, ranging from simple molecules like ammonia \( \text{NH}_3 \) to complex proteins in living organisms.
Understanding Oxidation States
Oxidation states, sometimes referred to as oxidation numbers, are a useful way to keep track of electrons in chemical reactions, especially in redox (reduction-oxidation) processes. These states indicate the degree of oxidation (loss of electrons) of an atom within a compound. The oxidation state is represented by an integer, which can be positive, negative, or zero.
  • Negative Oxidation State: Indicates gain of electrons.
  • Positive Oxidation State: Shows loss of electrons.
  • Zero Oxidation State: Element in its pure form.
For nitrogen, possible oxidation states include -3, +3, and +5, with -3 being the most stable state due to its ability to gain three electrons to imitate the electron configuration of noble gases. The different states of nitrogen allow it to form various compounds with distinct properties. For example, in \( \text{NH}_3 \), nitrogen exhibits an oxidation state of -3, while in \( \text{HNO}_3 \) (nitric acid), its oxidation state is +5.
Periodic Table Concepts
The Periodic Table is a tabular display of all known chemical elements, organized by increasing atomic number, electronic configuration, and recurring chemical properties. Its arrangement reveals the periodic law, where elements with similar properties recur at intervals. The groups, or vertical columns, represent elements with similar valence electron configurations, which explains their chemical behavior. For instance, the elements in Group 15, like nitrogen, often have five valence electrons and show varied oxidation states due to the possibility of gaining or sharing electrons to achieve a stable electronic configuration. Understanding periodic table concepts offers insights into the chemical reactivity and bonding characteristics of an element:
  • Atomic Number: Defines number of protons/electrons in an atom.
  • Groups: Vertical columns with elements sharing similar properties.
  • Periods: Horizontal rows indicating energy levels of electrons.
These concepts are foundational in predicting how elements, like nitrogen, interact in compounds and undergo chemical transformations. Knowing the group and period of an element allows us to infer its potential valency and behavior in chemical reactions.

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Most popular questions from this chapter

The increasing order of the first ionization enthalpies of the element \(\mathrm{B}, \mathrm{P}, \mathrm{S}\) and \(\mathrm{F}\) (lowest first) is: (a) \(\mathrm{F}<\mathrm{S}<\mathrm{P}<\mathrm{B}\) (b) \(\mathrm{P}<\mathrm{S}<\mathrm{B}<\mathrm{F}\) (c) \(\mathrm{B}<\mathrm{P}<\mathrm{S}<\mathrm{F}\) (d) \(\mathrm{B}<\mathrm{S} \leq \mathrm{P}<\mathrm{F}\)

Which of the following statements are correct? (1) HF is a stronger acid than \(\mathrm{HCl}\). (2) Among halide ions, iodide is the most powerful reducing agent. (3) Fluorine is the only halogen that does not show a variable oxidation state. (4) HOCl is a stronger acid than HOBr. (a) 2 and 4 (b) 2 and 3 (c) 1,2 and 3 (d) 2,3 and 4

Identify the correct order in which the ionic radius of the following ions increases: (1) F \(\begin{array}{ll}\text { (2) } \mathrm{Na}^{+} & \text {(3) } \mathrm{N}^{3-}\end{array}\) (a) \(3,1,2\) (b) \(1,2,3\) (c) \(2,3,1\) (d) \(2,1,3\)

The correct statement among the following is: (a) The first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\). (b) The second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of \(\mathrm{Na}\). (c) The first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\). (d) The third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of Al.

The electronic configuration of four elements are (1) \([\mathrm{Xe}] 6 \mathrm{~s}^{1}\) (2) \([\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\) (3) \([\mathrm{Ar}] 4 \mathrm{~s}^{2} 4 \mathrm{p}^{5}\) (4) \([\mathrm{Ar}] 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{2}\) Which one of the following statements about these elements is not correct? (a) (1) is a strong reducing agent (b) (2) is a d block element (c) (3) has high electron affinity (d) (4) shows variable oxidation state

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