Problem 120
Question
According to Modern Periodic Law, the physical and chemical properties of elements are a periodic function of their atomic number. When the elements are arranged according to this law, a table is obtained in which the elements with similar properties are placed together. The properties of elements are gradually changing but repeated at regular intervals. An element has exceptional outer electronic configuration as \(4 \mathrm{~d}^{10} 5 \mathrm{~s}^{0}\). It belongs to (a) 5 th period, group 10 (b) 4th period, group 11 (c) 5 th period, group 11 (d) 4th period, group 10
Step-by-Step Solution
Verified Answer
(a) 5th period, group 10.
1Step 1: Identify the Electron Configuration
The element has the outer electronic configuration of \(4d^{10} 5s^0\). This configuration indicates that the element has completed its 4d orbital and does not have electrons in the 5s orbital.
2Step 2: Determine the Period
The period number is traditionally given by the highest principal quantum number (n) that has electrons. In this case, the highest principal quantum number is 5 from the \(5s^0\) orbital, so the element is in the 5th period.
3Step 3: Determine the Group Number
For elements in the d-block (transition metals), the group number can be calculated by adding the number of electrons in the ns orbital and the (n-1)d orbitals together. Here, there are 0 electrons in the 5s and 10 electrons in the 4d. Therefore, the group number is \(0 + 10 = 10\).
4Step 4: Identify the Element and Match the Option
With the electronic configuration \(4d^{10} 5s^0\), the element is Palladium (Pd), which is known to belong to the 5th period and group 10. Therefore, the correct option is (a) 5th period, group 10.
Key Concepts
Electron ConfigurationTransition MetalsPeriodic TableAtomic Number
Electron Configuration
Electron configuration is like an address for electrons in an atom. It tells us where each electron resides. To write it, we express the distribution of electrons in atomic orbitals – such as s, p, d, and f. Each letter represents a different type of orbital with a unique shape. In our scenario, we deal with a configuration of \(4d^{10} 5s^0\). This means:
- 4d has the maximum number of electrons (10).
- 5s is empty (0 electrons).
Transition Metals
Transition metals are a special group of elements found in the d-block of the periodic table. They are known for their vibrant colors and versatile chemical behavior. A classic feature of these elements is their ability to form multiple oxidation states due to specific electron configurations.
- They have partially filled d orbitals.
- Exhibit characteristic properties like magnetism and conductivity.
- Include elements like iron, copper, and palladium (our element in question).
Periodic Table
The periodic table is like a map that organizes elements based on their atomic number, electron configuration, and recurring chemical properties. It's divided into blocks and groups:
- Blocks: Spanning the s, p, d, and f areas.
- Groups: Elements in the same column share similarities.
Atomic Number
An atom's atomic number is fundamental – it defines an element by telling us the number of protons in its nucleus and, typically, the number of electrons in a neutral atom. For Palladium (Pd), the atomic number is 46.
- This number is crucial for placing elements in the periodic table.
- It influences the chemical and physical properties of the element.
Other exercises in this chapter
Problem 117
Properties that are directly or indirectly related to the electronic configuration of the elements and show a regular gradation when we move from left to right
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Properties that are directly or indirectly related to the electronic configuration of the elements and show a regular gradation when we move from left to right
View solution Problem 121
According to Modern Periodic Law, the physical and chemical properties of elements are a periodic function of their atomic number. When the elements are arrange
View solution Problem 123
Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{ll} \text { (a) } \mathrm{C}>\mathrm{N} & \text
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