Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Transition element exhibits variable oxidation states because they release electrons from the following orbitals. (1) ns and np orbitals (2) \((n-1) \mathrm{d}\) and ns orbitals (3) \((n-1)\) d orbital (4) ns orbital

Short Answer

Expert verified
Option (2) \((n-1)\text{d}\) and ns orbitals.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Analyze the Question

Identify the correct orbitals from which transition elements release electrons to exhibit variable oxidation states.
02

Understanding Transition Elements

Transition elements are known for their ability to exhibit multiple oxidation states. This property is primarily due to the involvement of electrons from their d-orbitals and s-orbitals.
03

Identify Applicable Orbitals

In transition elements, electrons are removed first from the (n-1)d orbital and then from the ns orbital. This is because the energy difference between these orbitals is small.
04

Eliminate Incorrect Options

Consider the given options: (1) ns and np orbitals (2) \((n-1)\text{d}\) and ns orbitals (3) \((n-1)\text{d}\) orbital (4) ns orbital. Options (1), (3), and (4) are not correct because they do not include both the (n-1)d and ns orbitals which are needed for variable oxidation states.
05

Select the Correct Option

The correct orbitals for the transition elements that exhibit variable oxidation states are \((n-1)\text{d}\) and ns orbitals, which corresponds to option (2).

Key Concepts

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

Variable Oxidation States
Transition elements are unique because they can exhibit a variety of oxidation states. This means they can lose different numbers of electrons to form ions. One reason for this is that their electron configurations allow for versatile bonding. When a transition metal forms a compound, it can lose electrons from both the s-orbital and d-orbital. This variability in electron loss is why transition elements like iron can have oxidation states of +2 or +3 and manganese can have oxidation states ranging from +2 to +7. These multiple oxidation states are crucial in many chemical reactions and industrial processes.
d-Orbitals
One of the key features of transition elements is the presence of d-orbitals. The d-orbitals are a set of five orbitals found in the third energy level of an atom and beyond (starting from the fourth period of the periodic table). These orbitals can hold up to 10 electrons. In transition elements, the d-orbital plays a significant role because it is not completely filled. This incomplete filling is what gives rise to complex electronic structures and the ability to form various oxidation states. For instance, electrons are first removed from the 4s orbital before the 3d orbital during ionization. The nearness in energy levels of these orbitals allows for this flexibility.
s-Orbitals
s-Orbitals are simpler in structure compared to d-orbitals. Each s-orbital can hold a maximum of 2 electrons. In transition metals, the s-orbital is generally filled before the d-orbital when you write the electron configuration. However, upon ionization, electrons are removed from the s-orbital first. For instance, the electron configuration of iron (Fe) is [Ar] 4s^2 3d^6. During the formation of Fe^2+ ion, the electrons are first removed from the 4s orbital, leading to the configuration of [Ar] 3d^6. Understanding the behavior of s-orbitals is essential to grasp why transition elements exhibit variable oxidation states.
Electron Configuration
Electron configuration refers to the arrangement of electrons in the orbitals of an atom. For transition elements, this configuration includes both the s-orbital and the d-orbital. Typically, an electron configuration might look like [noble gas] ns^2 (n-1)d^x, where n is the period number and x is the number of electrons in the d-orbital. For example, the electron configuration of cobalt (Co) is [Ar] 4s^2 3d^7. Such configurations explain why transition metals can lose different numbers of electrons from both the s-orbital and the d-orbital, leading to variable oxidation states. Understanding electron configurations can help predict the chemical and physical properties of the elements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The false statement among the following is (1) Electron affinity of noble gases is almost zero. (2) The halogen with highest electron affinity is fluorine. (3) Electron affinity values are obtained indirectly by Born-Haber Cycle. (4) lonisation potential of Na would be numerically the same as electron affinity of \(\mathrm{Na}^{+}\).

The position of the clement in the periodic table with its outer electronic configuration \(3 \mathrm{~s}^{2} 3 \mathrm{p}^{4}\) is (1) IV period, VIB group (2) IIl period, VIA group (3) IV period, IVB group (4) IIl period, IVB group

Which among the following statements is false? (1) Pauling electronegativity scale is based on the experimental value of bond energies. (2) Electronegativity is a measure of the capacity of an atom to attract shared pair of electrons. (3) The atom with high electronegativity generally has high ionisation potential. (4) The electronegativity difference between the atoms gives an idea about bond length.

Which of the series of elements listed below would have nearly same atomic radii (1) \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}\) (2) \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}\) (3) \(\mathrm{Li}, \mathrm{Be}, \mathrm{B}, \mathrm{C}\) (4) \(\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni}, \mathrm{Cu}\)

The correct order of sccond ionisation potential of \(\mathrm{C}, \mathrm{N}, \mathrm{O}\) and \(\mathrm{F}\) is (1) \(\mathrm{C}>\mathrm{N}>\mathrm{O}>\mathrm{F}\) (2) \(\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}\) (3) \(\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}\) (4) \(\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free