
Concept explainers
Group the species that are isoelectronic:

Interpretation:
Theisoelectronic species are to be identified from the given species.
Concept introduction:
The electronic configuration of a noble gas is
The species that have identical electronic configurations but different nuclear charges are called isoelectronic species.
Answer to Problem 59QP
Solution: The ion-pairs are
Explanation of Solution
A neutral atom in its ground state cannot be isoelectric with any other element. However, ions may have the same electronic configurations with different atoms or ions.
The electronic configurations of the given elements are as follows:
Electronic configuration for beryllium ion is as follows:
There are total two electrons in beryllium ion.
Electronic configuration for helium atom is as follows:
There are total two electrons in helium atom.
The electronic configurations of
Electronic configuration for fluorine ion is as follows:
There are total 10electrons in fluorine ion.
Electronic configuration for nitrogen ion is as follows:
There are total 10electrons in nitrogen ion.
The electronic configurations of
Electronic configuration for sulphur ion is as follows:
There are total 18 electrons in sulphur ion.
Electronic configuration for argon atom is as follows:
There are total 18electrons in argon atom.
The electronic configurations of
Electronic configuration for iron ion is as follows:
There are total24 electrons in iron ion.
Electronic configuration for cobalt ion is as follows:
There are total 24electrons in cobalt ion.
The electronic configurations of
Hence, the ion-pairs
The ion-pairs
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Chapter 7 Solutions
Chemistry
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:arrow_forward3:48 PM Fri Apr 4 K Problem 4 of 10 Submit Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Mg. :0: Select to Add Arrows :0: :Br: Mg :0: :0: Select to Add Arrows Mg. Br: :0: 0:0- Br -190 H 0:0 Select to Add Arrows Select to Add Arrows neutralizing workup H CH3arrow_forwardIarrow_forward
- Draw the Markovnikov product of the hydrobromination of this alkene. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for this problem. + Explanation Check 1 X E 4 1 1 1 1 1 HBr Click and drag to start drawing a structure. 80 LE #3 @ 2 $4 0 I அ2 % 85 F * K M ? BH 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center & 6 27 FG F10 8 9 R T Y U D F G H P J K L Z X C V B N M Q W A S H option command H command optiarrow_forwardBe sure to use wedge and dash bonds to show the stereochemistry of the products when it's important, for example to distinguish between two different major products. Predict the major products of the following reaction. Explanation Q F1 A Check F2 @ 2 # 3 + X 80 F3 W E S D $ 4 I O H. H₂ 2 R Pt % 05 LL ee F6 F5 T <6 G Click and drag to start drawing a structure. 27 & A 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Acce Y U H DII 8 9 F10 4 J K L Z X C V B N M T H option command F11 P H commandarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and product in this reaction or mechanistic step(s). Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore inorganic byproducts. H :0: CH3 O: OH Q CH3OH2+ Draw Intermediate protonation CH3OH CH3OH nucleophilic addition H Draw Intermediate deprotonation :0: H3C CH3OH2* protonation H 0: H CH3 H.arrow_forward
- Predicting the reactants or products of hemiacetal and acetal formation uentify the missing organic reactants in the following reaction: H+ X+Y OH H+ за Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H2O) are not shown. In the drawing area below, draw the skeletal ("line") structures of the missing organic reactants X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Explanation Check Click and drag to start drawing a structure. ? olo 18 Ar © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibilityarrow_forwardcan someone please answer thisarrow_forwardPlease, please help me figure out the the moles, molarity and Ksp column. Step by step details because I've came up with about three different number and have no idea what I'm doing wrong.arrow_forward
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